双指针

Author: luzhipeng

19.remove-nth-node-from-end-of-list.js

@@ -0,0 +1,50 @@
+/*
+ * @lc app=leetcode id=19 lang=javascript
+ *
+ * [19] Remove Nth Node From End of List
+ *
+ * https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/
+ *
+ * algorithms
+ * Medium (34.03%)
+ * Total Accepted:    360.1K
+ * Total Submissions: 1.1M
+ * Testcase Example:  '[1,2,3,4,5]\n2'
+ *
+ * Given a linked list, remove the n-th node from the end of list and return
+ * its head.
+ * 
+ * Example:
+ * 
+ * 
+ * Given linked list: 1->2->3->4->5, and n = 2.
+ * 
+ * After removing the second node from the end, the linked list becomes
+ * 1->2->3->5.
+ * 
+ * 
+ * Note:
+ * 
+ * Given n will always be valid.
+ * 
+ * Follow up:
+ * 
+ * Could you do this in one pass?
+ * 
+ */
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @param {number} n
+ * @return {ListNode}
+ */
+var removeNthFromEnd = function(head, n) {
+    
+};
+

assets/19.removeNthNodeFromEndOfList.gif

@@ -0,0 +1,131 @@
+## 题目地址
+https://leetcode.com/problems/remove-nth-node-from-end-of-list/description
+
+## 题目描述
+Given a linked list, remove the n-th node from the end of list and return its head.
+
+Example:
+
+Given linked list: 1->2->3->4->5, and n = 2.
+
+After removing the second node from the end, the linked list becomes 1->2->3->5.
+Note:
+
+Given n will always be valid.
+
+Follow up:
+
+Could you do this in one pass?
+
+## 思路
+
+双指针，指针A先移动n次， 指针B再开始移动。当A到达null的时候， 指针b的位置正好是倒数n
+
+我们可以设想假设设定了双指针p和q的话，当q指向末尾的NULL，p与q之间相隔的元素个数为n时，那么删除掉p的下一个指针就完成了要求。
+
+设置虚拟节点dummyHead指向head
+
+设定双指针p和q，初始都指向虚拟节点dummyHead
+
+移动q，直到p与q之间相隔的元素个数为n
+
+同时移动p与q，直到q指向的为NULL
+
+将p的下一个节点指向下下个节点
+
+
+
+![19.removeNthNodeFromEndOfList](./assets/19.removeNthNodeFromEndOfList.gif)
+
+(图片来自： https://github.com/MisterBooo/LeetCodeAnimation)
+
+## 关键点解析
+
+1. 链表这种数据结构的特点和使用
+
+2. 使用双指针
+
+3. 使用一个dummyHead简化操作
+
+## 代码
+
+
+```js
+/*
+ * @lc app=leetcode id=19 lang=javascript
+ *
+ * [19] Remove Nth Node From End of List
+ *
+ * https://leetcode.com/problems/remove-nth-node-from-end-of-list/description/
+ *
+ * algorithms
+ * Medium (34.03%)
+ * Total Accepted:    360.1K
+ * Total Submissions: 1.1M
+ * Testcase Example:  '[1,2,3,4,5]\n2'
+ *
+ * Given a linked list, remove the n-th node from the end of list and return
+ * its head.
+ * 
+ * Example:
+ * 
+ * 
+ * Given linked list: 1->2->3->4->5, and n = 2.
+ * 
+ * After removing the second node from the end, the linked list becomes
+ * 1->2->3->5.
+ * 
+ * 
+ * Note:
+ * 
+ * Given n will always be valid.
+ * 
+ * Follow up:
+ * 
+ * Could you do this in one pass?
+ * 
+ */
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @param {number} n
+ * @return {ListNode}
+ */
+var removeNthFromEnd = function(head, n) {
+  let i = -1;
+  const noop = {
+    next: null
+  };
+
+  const dummyHead = new ListNode(); // 增加一个dummyHead 简化操作
+  dummyHead.next = head;
+
+  let currentP1 = dummyHead;
+  let currentP2 = dummyHead;
+
+  
+  while (currentP1) {
+
+    if (i === n) {
+      currentP2 = currentP2.next;
+    }
+
+    if (i !== n) {
+        i++;
+    }
+    
+    currentP1 = currentP1.next;
+  }
+
+  currentP2.next = ((currentP2 || noop).next || noop).next;
+
+  return dummyHead.next;
+};
+
+```