237 Delete Node in a Linked List

Author: int32bit

README.md

@@ -129,6 +129,7 @@
 + [230 Kth Smallest Element in a BST(BST、中序遍历)](algorithms/KthSmallestElementinaBST)
 + [231 Power of Two(位运算，二进制1的个数)](algorithms/PowerofTwo)
 + [232 Implement Queue using Stacks(栈模拟队列)](algorithms/ImplementQueueusingStacks)
++ [237 Delete Node in a Linked List(O(1)删除单链表节点)](algorithms/DeleteNodeinaLinkedList)
 
 ## Database
 

algorithms/DeleteNodeinaLinkedList/README.md

@@ -0,0 +1,22 @@
+## Delete Node in a Linked List 
+
+Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.
+
+Supposed the linked list is `1 -> 2 -> 3 -> 4` and you are given the third node with value `3`, the linked list should become `1 -> 2 -> 4` after calling your function.
+
+## Solution
+
+给定一个单链表节点，删除该节点。
+
+由于无法获得该节点的前驱节点，因此无法更新前驱节点的后继节点。我们可以把该节点的下一个节点的值拷贝到当前节点，转而删除该节点的下一个节点。
+
+该题目也可以描述为: 如何在O(1)的时间删除单链表的某个节点！
+
+```cpp
+void deleteNode(ListNode *node) {
+	ListNode *p = node->next;
+	node->val = p->val;
+	node->next = p->next;
+	delete p;
+}
+```

algorithms/DeleteNodeinaLinkedList/list.cpp

@@ -0,0 +1,77 @@
+#include <iostream>
+#include <cstdio>
+#include <vector>
+#include <algorithm>
+#include <string>
+using namespace std;
+#include <stdlib.h>
+struct ListNode {
+	int val;
+	ListNode *next;
+	ListNode(int x): val(x), next(nullptr) {}
+};
+class Solution {
+	public:
+		void deleteNode(ListNode *node) {
+			ListNode *p = node->next;
+			node->val = p->val;
+			node->next = p->next;
+			delete p;
+		}
+
+};
+int getLength(ListNode *head)
+{
+	int len = 0;
+	ListNode *p = head;
+	while (p) {
+		++len;
+		p = p->next;
+	}
+	return len;
+}
+void print(ListNode *head)
+{
+	if (head == nullptr) {
+		printf("NULL\n");
+		return;
+	}
+	struct ListNode *p = head;
+	while (p) {
+		printf("%d ", p->val);
+		p = p->next;
+	}
+	printf("\n");
+}
+ListNode * mk_list(ListNode **ha, int a[], int n)
+{
+	if (n < 1)
+		return nullptr;
+	ListNode *p = new ListNode(a[0]);
+	*ha = p;
+	for (int i = 1; i < n; ++i) {
+		ListNode *q = new ListNode(a[i]);
+		p->next = q;
+		p = q;
+	}
+	return p;
+}
+void free_list(struct ListNode *head)
+{
+	struct ListNode *p = head;
+	while (p) {
+		struct ListNode *q = p->next;
+		delete p;
+		p = q;
+	}
+}
+int main(int argc, char **argv)
+{
+	Solution solution;
+	struct ListNode *head = NULL;
+	int a[] = {1, 2, 3};
+	mk_list(&head, a, 3);
+	solution.deleteNode(head->next);
+	print(head);
+	return 0;
+}