20190818

Author: mJackie

.gitignore

@@ -1,4 +1,5 @@
 .DS_Store
 */.DS_Store
-/kickstart/*
+kickstart/*
 test.java
+interview/*

code/lc1026.java

@@ -1,6 +1,6 @@
 package code;
 /*
- * 1025. Maximum Difference Between Node and Ancestor
+ * 1026. Maximum Difference Between Node and Ancestor
  * 题意：父节点减子节点的绝对值最大
  * 难度：
  * 分类：

code/lc105.java

@@ -37,4 +37,17 @@ public static TreeNode recursion(int[] preorder, int[] inorder, int pre_index, i
         return tn;      //记住函数的返回值的设置，返回Node，递归的构造子树
     }
 
+    public TreeNode buildTree2(int[] preorder, int[] inorder) {
+        return helper(preorder, inorder, 0, 0, preorder.length-1);
+    }
+    public TreeNode helper(int[] preorder, int[] inorder, int cur, int left, int right){
+        if(left>right) return null;
+        TreeNode tn = new TreeNode(preorder[cur]);
+        int i = left;
+        for(; i<=right; i++) if(inorder[i]==preorder[cur]) break;
+        tn.left = helper(preorder, inorder, cur+1, left, i-1);
+        tn.right = helper(preorder, inorder, cur+i-left+1, i+1, right);
+        return tn;
+    }
+
 }

code/lc113.java

@@ -33,7 +33,7 @@ public void helper(List<List<Integer>> res, List<Integer> cur, TreeNode root, in
             helper(res, cur, root.left, sum-root.val);
             helper(res, cur, root.right, sum-root.val);
         }
-        cur.remove(cur.size()-1);
+        cur.remove(cur.size()-1);   //注意是去掉最后一个，传的是索引。传递对象的话，序列可能会变。
         return;
     }
 }

code/lc148.java

@@ -75,9 +75,9 @@ public void quickSort(ListNode head, ListNode end) {
 
     public ListNode partion(ListNode head, ListNode end) {
         ListNode p1 = head, p2 = head.next;
-
+        //p1指的是小于pivot的索引
         //走到末尾才停
-        while (p2 != end) { //p1与p2间都是大于pivot的数
+        while (p2 != end) { //head到p1间是小于pivot的数，p1与p2间都是大于pivot的数
             if (p2.val < head.val) {    //lc922 类似的思想， 把小于的值放到该放的位置上
                 p1 = p1.next;
 

code/lc149.java

@@ -20,7 +20,7 @@ class Point {
     public int maxPoints(Point[] points) {
         if(points.length<=2) return points.length;
         int res = 0;
-        for (int i = 0; i < points.length-1 ; i++) {
+        for (int i = 0; i < points.length-1 ; i++) {//先固定一个点，再根据斜率进行记录就行了
             HashMap<Integer, HashMap<Integer, Integer>> hm = new HashMap<>();
             int overlap = 1;
             int max = 0;

code/lc153.java

@@ -16,7 +16,7 @@ public int findMin(int[] nums) {
             int mid = (left+right)/2;
             if(nums[mid]<nums[right]){
                 right = mid;    //这不加1
-            }else if(nums[mid]>nums[right]){
+            }else{
                 left = mid+1;
             }
         }

code/lc22.java

@@ -1,7 +1,7 @@
 package code;
 /*
  * 22. Generate Parentheses
- * 题意：正确括号组合的
+ * 题意：正确括号组合
  * 难度：Medium
  * 分类：String, Backtracking
  * 思路：回溯法的典型题目，按选优条件向前搜索，达到目标后就退回一步或返回

code/lc236.java

@@ -23,10 +23,8 @@ public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {//
             return root;
         TreeNode left = lowestCommonAncestor(root.left, p, q);
         TreeNode right = lowestCommonAncestor(root.right, p, q);
-        if( left!=null && right!=null )     //回溯返回。哪边不为空，返回哪边，否则返回自己。
-            return root;
-        else if(left!=null)
-            return left;
+        if( left!=null && right!=null ) return root;     //回溯返回。哪边不为空，返回哪边，否则返回自己。
+        else if(left!=null) return left;
         else return right;
     }
 

code/lc239.java

@@ -25,7 +25,7 @@ public static int[] maxSlidingWindow(int[] nums, int k) {
             return new int[]{};
         int[] res = new int[nums.length-k+1];
         int cur = 0;
-        Deque<Integer> dq = new ArrayDeque();   //队列里是递减的
+        Deque<Integer> dq = new ArrayDeque();   //队列里是递减的，存的仍然是下标
         for (int i = 0; i < nums.length ; i++) {
             if( !dq.isEmpty() && dq.peekFirst()<=i-k)   //窗口长度过长了，删掉头
                 dq.removeFirst();