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‎README.md‎

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<h1 align="center">algorithm-tutorial</h1>
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> 算法、数据结构这类知识点真的需要天分。本人不是算法工程师，也做不了算法工程师。总是在见识到一些算法大神的优雅实现后，感叹于自己是多么的弱智，同样是写代码，我写的是什么垃圾！
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> 算法、数据结构学习要点：
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>
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> 这个项目的初衷，是将工作中经常会接触到的一些数据结构和算法整理归纳。人总是要有理想的，要想少写垃圾的代码，必须夯实算法的基础。路漫漫兮其修远，吾将上下而求索。
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> 三分学，七分练
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>
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> 坚持 + 坚持 + 坚持
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>
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> 🔁 项目同步维护在 [github](https://github.com/dunwu/algorithm-tutorial) | [gitee](https://gitee.com/turnon/algorithm-tutorial)
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>
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> 📖 [电子书](https://dunwu.github.io/algorithm-tutorial/) | [电子书（国内）](http://turnon.gitee.io/algorithm-tutorial/)
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## 内容
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![img](http://dunwu.test.upcdn.net/snap/20200702071922.png)
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- [算法概述](docs/overview.md)
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- [数组](docs/array.md)
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- [链表](docs/list.md)
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- [栈](docs/stack.md)
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#### 二叉树
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- [二叉搜索树的最近公共祖先](https://github.com/dunwu/algorithm-tutorial/blob/master/codes/algorithm/src/main/java/io/github/dunwu/algorithm/tree/btree/二叉搜索树的最近公共祖先.java)
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- [二叉搜索树节点最小距离](https://github.com/dunwu/algorithm-tutorial/blob/master/codes/algorithm/src/main/java/io/github/dunwu/algorithm/tree/btree/二叉搜索树节点最小距离.java)
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- [二叉树的层次遍历](https://github.com/dunwu/algorithm-tutorial/blob/master/codes/algorithm/src/main/java/io/github/dunwu/algorithm/tree/btree/二叉树的层次遍历.java)
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- [二叉树的层次遍历 2](https://github.com/dunwu/algorithm-tutorial/blob/master/codes/algorithm/src/main/java/io/github/dunwu/algorithm/tree/btree/二叉树的层次遍历2.java)

‎assets/算法.xmind‎

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‎codes/algorithm/src/main/java/io/github/dunwu/algorithm/divide/N次幂.java‎

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package io.github.dunwu.algorithm.divide;
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import org.junit.jupiter.api.Assertions;
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/**
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* 50. Pow(x, n) 求 N次幂
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*
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* @author <a href="mailto:forbreak@163.com">Zhang Peng</a>
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* @see <a href="https://leetcode-cn.com/problems/powx-n/">50. Pow(x, n)</a>
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* @since 2020-07-02
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*/
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public class N次幂 {
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public static void main(String[] args) {
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Assertions.assertEquals(1024.00000, myPow(2.00000, 10));
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Assertions.assertEquals(9.261000000000001, myPow(2.10000, 3));
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Assertions.assertEquals(0.25000, myPow(2.00000, -2));
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Assertions.assertEquals(1024.00000, myPow2(2.00000, 10));
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Assertions.assertEquals(9.261000000000001, myPow2(2.10000, 3));
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Assertions.assertEquals(0.25000, myPow2(2.00000, -2));
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}
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/**
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* 分治算法
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* <p>
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* x^n 可以视为 y^2(n为偶数) 或 x*y^2(n为奇数)
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* <p>
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* 时间复杂度：O(log N)
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*/
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public static double myPow(double x, int n) {
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if (n > 0) return helper(x, n);
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return 1.00000 / helper(x, -n);
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}
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public static double helper(double x, int n) {
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if (n == 0) {
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return 1.00000;
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}
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double y = helper(x, n / 2);
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if (n % 2 == 0) {
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return y * y;
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} else {
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return x * y * y;
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}
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}
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/**
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* 暴力破解法
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* <p>
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* 时间复杂度：O(N)
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*/
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public static double myPow2(double x, int n) {
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double result = 1.00000;
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if (n == 0) {
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return 1.00000;
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}
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int cnt = n > 0 ? n : -n;
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for (int i = 0; i < cnt; i++) {
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result *= x;
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}
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if (n < 0) result = 1.00000 / result;
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return result;
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}
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// 库函数
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public static double myPow3(double x, int n) {
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return Math.pow(x, n);
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}
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}

‎codes/algorithm/src/main/java/io/github/dunwu/algorithm/divide/package-info.java‎

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/**
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* 分治算法
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*
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* @author <a href="mailto:forbreak@163.com">Zhang Peng</a>
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* @since 2020-07-02
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*/
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package io.github.dunwu.algorithm.divide;

‎codes/algorithm/src/main/java/io/github/dunwu/algorithm/divide/多数元素.java‎

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package io.github.dunwu.algorithm.divide;
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import org.junit.jupiter.api.Assertions;
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import java.util.Arrays;
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/**
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* 多数元素
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*
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* @author <a href="mailto:forbreak@163.com">Zhang Peng</a>
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* @see <a href="https://leetcode-cn.com/problems/majority-element/">多数元素</a>
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* @since 2020-07-02
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*/
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public class 多数元素 {
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public static void main(String[] args) {
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Assertions.assertEquals(3, majorityElement(new int[] { 3, 2, 3 }));
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Assertions.assertEquals(2, majorityElement(new int[] { 2, 2, 1, 1, 1, 2, 2 }));
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Assertions.assertEquals(6, majorityElement(new int[] { 6, 6, 6, 7, 7 }));
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}
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// 暴力破解法
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// 时间复杂度：O(N) + O(log N)
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public static int majorityElement(int[] nums) {
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Arrays.sort(nums);
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int max = 1;
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int count = 0;
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int currElem = nums[0];
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int maxElem = nums[0];
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for (int i = 0; i < nums.length; i++) {
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if (nums[i] != currElem) {
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count = 1;
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currElem = nums[i];
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} else {
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count++;
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if (maxElem == currElem) {
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max = count;
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} else {
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if (max < count) {
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maxElem = currElem;
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}
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}
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}
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}
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return maxElem;
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}
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}

‎codes/algorithm/src/main/java/io/github/dunwu/algorithm/tree/TreeUtils.java‎

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}
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}
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public static TreeNode find(TreeNode root, int val) {
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if (root == null || root.val == val) { return root;}
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TreeNode left = find(root.left, val);
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if (left != null) return left;
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return find(root.right, val);
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}
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public static void depthOrderTraverse(TreeNode root) {
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if (root == null) {
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return;

‎codes/algorithm/src/main/java/io/github/dunwu/algorithm/tree/btree/二叉搜索树的最近公共祖先.java‎

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package io.github.dunwu.algorithm.tree.btree;
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import io.github.dunwu.algorithm.tree.TreeUtils;
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import org.junit.jupiter.api.Assertions;
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/**
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* <code>235. 二叉搜索树的最近公共祖先</code> 算法实现
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*
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* @see 二叉树的最近公共祖先 可以使用二叉树的最近公共祖先，但没有利用二叉搜索树特性，性能略差
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* @see <a href="https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-search-tree/">235. 二叉搜索树的最近公共祖先</a>
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*/
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public class 二叉搜索树的最近公共祖先 {
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public static void main(String[] args) {
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TreeNode root = TreeUtils.buildTree(6, 2, 8, 0, 4, 7, 9, null, null, 3, 5);
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TreeNode p = TreeUtils.find(root, 2);
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TreeNode q = TreeUtils.find(root, 8);
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// TreeNode treeNode = lowestCommonAncestor(root, p, q);
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TreeNode treeNode = lowestCommonAncestor2(root, p, q);
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Assertions.assertNotNull(treeNode);
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Assertions.assertEquals(6, treeNode.val);
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System.out.println("公共祖先节点 = " + treeNode.val);
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TreeNode root2 = TreeUtils.buildTree(6, 2, 8, 0, 4, 7, 9, null, null, 3, 5);
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TreeNode p2 = TreeUtils.find(root2, 2);
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TreeNode q2 = TreeUtils.find(root2, 4);
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// TreeNode treeNode2 = lowestCommonAncestor(root2, p2, q2);
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TreeNode treeNode2 = lowestCommonAncestor2(root2, p2, q2);
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Assertions.assertNotNull(treeNode2);
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Assertions.assertEquals(2, treeNode2.val);
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System.out.println("公共祖先节点 = " + treeNode2.val);
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}
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/**
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* 递归方式求解
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* <p>
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* 时间复杂度：O(N) 线性级
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* <p>
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* 空间复杂度：O(2) 常数级
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*/
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public static TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
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// 如果当前节点为空，直接返回
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// 或当前节点就是 p 或 q 其中一个，显然就是要找的最近公共祖先，直接返回
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if (root == null || root == p || root == q) return root;
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if (root.val > p.val && root.val > q.val) { // 如果当前节点值同时大于 p、q 的值，说明 p、q 肯定都在左子树
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return lowestCommonAncestor(root.left, p, q);
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} else if (root.val < p.val && root.val < q.val) { // 如果当前节点值同时小于 p、q 的值，说明 p、q 肯定都在右子树
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return lowestCommonAncestor(root.right, p, q);
50+
} else {
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return root;
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}
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}
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/**
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* 非递归方式求解
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* <p>
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* 时间复杂度：O(N) 线性级
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* <p>
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* 空间复杂度：O(2) 常数级
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*/
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public static TreeNode lowestCommonAncestor2(TreeNode root, TreeNode p, TreeNode q) {
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// 如果当前节点为空，直接返回
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// 或当前节点就是 p 或 q 其中一个，显然就是要找的最近公共祖先，直接返回
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if (root == null || root == p || root == q) return root;
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TreeNode curr = root;
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while (curr != null) {
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if (curr.val > p.val && curr.val > q.val) { // 如果当前节点值同时大于 p、q 的值，说明 p、q 肯定都在左子树
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curr = curr.left;
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} else if (curr.val < p.val && curr.val < q.val) { // 如果当前节点值同时小于 p、q 的值，说明 p、q 肯定都在右子树
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curr = curr.right;
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} else {
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return curr;
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}
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}
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return curr;
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}
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}

‎codes/algorithm/src/main/java/io/github/dunwu/algorithm/tree/btree/二叉树的最近公共祖先.java‎

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package io.github.dunwu.algorithm.tree.btree;
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import io.github.dunwu.algorithm.tree.TreeUtils;
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import io.github.dunwu.algorithm.tree.btree.TreeNode;
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import org.junit.jupiter.api.Assertions;
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/**
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* <code>236. 二叉树的最近公共祖先</code> 算法实现
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* <p>
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* 给定一个二叉树, 找到该树中两个指定节点的最近公共祖先。
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* <p>
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* 百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”
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* <p>
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* 例如，给定如下二叉树: root = [3,5,1,6,2,0,8,null,null,7,4]
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* <p>
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* 示例 1:
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* <pre>
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* 输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
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* 输出: 3
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* 解释: 节点 5 和节点 1 的最近公共祖先是节点 3。
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* </pre>
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* 示例 2:
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* <pre>
23-
* 输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
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* 输出: 5
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* 解释: 节点 5 和节点 4 的最近公共祖先是节点 5。因为根据定义最近公共祖先节点可以为节点本身。
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* </pre>
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* 说明:
28-
* <pre>
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* 所有节点的值都是唯一的。
30-
* p、q 为不同节点且均存在于给定的二叉树中。
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* </pre>
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*
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* @see <a href="https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-tree/">236. 二叉树的最近公共祖先</a>
10+
* @see <a href="https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-tree/solution/er-cha-shu-de-zui-jin-gong-gong-zu-xian-by-leetc-2/">解题思路</a>
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*/
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public class 二叉树的最近公共祖先 {
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37-
private static TreeNode ans = null;
38-
3914
public static void main(String[] args) {
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TreeNode root = TreeUtils.buildTree(3, 5, 1, 6, 2, 0, 8, null, null, 7, 4);
41-
TreeNode p = new TreeNode(5);
42-
TreeNode q = new TreeNode(1);
16+
TreeNode p = TreeUtils.find(root, 5);
17+
TreeNode q = TreeUtils.find(root, 1);
4318
TreeNode treeNode = lowestCommonAncestor(root, p, q);
19+
Assertions.assertNotNull(treeNode);
20+
Assertions.assertEquals(3, treeNode.val);
4421
System.out.println("公共祖先节点 = " + treeNode.val);
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23+
TreeNode p2 = TreeUtils.find(root, 5);
24+
TreeNode q2 = TreeUtils.find(root, 4);
25+
TreeNode treeNode2 = lowestCommonAncestor(root, p2, q2);
26+
Assertions.assertNotNull(treeNode2);
27+
Assertions.assertEquals(5, treeNode2.val);
28+
System.out.println("公共祖先节点 = " + treeNode2.val);
4529
}
4630

31+
/**
32+
* 递归方式求解
33+
* <p>
34+
* 时间复杂度：O(N) 线性级
35+
* <p>
36+
* 空间复杂度：O(2) 常数级
37+
*/
4738
public static TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
48-
recurseTree(root, p, q);
49-
return ans;
50-
}
39+
// 如果当前节点为空，直接返回
40+
// 或当前节点就是 p 或 q 其中一个，显然就是要找的最近公共祖先，直接返回
41+
if (root == null || root == p || root == q) return root;
42+
43+
TreeNode left = lowestCommonAncestor(root.left, p, q);
44+
TreeNode right = lowestCommonAncestor(root.right, p, q);
5145

52-
private static boolean recurseTree(TreeNode root, TreeNode p, TreeNode q) {
53-
if (root == null) return false;
54-
boolean lson = recurseTree(root.left, p, q);
55-
boolean rson = recurseTree(root.right, p, q);
56-
boolean flag = (root.val == p.val || root.val == q.val) && (lson || rson);
57-
if ((lson && rson) || ((root.val == p.val || root.val == q.val) && (lson || rson))) { ans = root; }
58-
return (root.val == p.val || root.val == q.val) || (lson || rson);
46+
if (left == null) { // p、q 都不在左子树，查找右子树
47+
return right;
48+
} else if (right == null) { // p、q 都不在右子树，查找左子树
49+
return left;
50+
} else {
51+
// p、q 既不在左子树，又不在右子树，直接返回当前节点
52+
return root;
53+
}
5954
}
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}

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