19. 删除链表的倒数第 N 个结点

Author: w2xi

README.md

@@ -19,6 +19,7 @@
 |9|[回文数](https://leetcode-cn.com/problems/palindrome-number/)|[JavaScript](./algorithms/palindrome-number.js)|Easy|
 |13|[罗马数字转整数](https://leetcode.cn/problems/roman-to-integer/)|[JavaScript](./algorithms/roman-to-integer.js)|Easy|
 |14|[最长公共前缀](https://leetcode-cn.com/problems/longest-common-prefix/)|[JavaScript](./algorithms/longest-common-prefix.js)|Easy|
+|19|[删除链表的倒数第 N 个结点](https://leetcode.cn/problems/remove-nth-node-from-end-of-list/)|[JavaScript](./algorithms/remove-nth-node-from-end-of-list.js)|Medium|
 |20|[有效的括号](https://leetcode.cn/problems/valid-parentheses/)|[JavaScript](./algorithms/valid-parentheses.js)|Easy|
 |21|[合并两个有序链表](https://leetcode-cn.com/problems/merge-two-sorted-lists/)|[JavaScript](./algorithms/merge-two-sorted-lists.js)|Easy|
 |24|[两两交换链表中的节点](https://leetcode.cn/problems/swap-nodes-in-pairs/)|[JavaScript](./algorithms/swap-nodes-in-pairs.js)|Medium|

algorithms/remove-nth-node-from-end-of-list.js

@@ -0,0 +1,88 @@
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @param {number} n
+ * @return {ListNode}
+ */
+var removeNthFromEnd = function (head, n) {
+  // 常规做法 计算链表长度
+  let dummyNode = new ListNode(-1);
+  dummyNode.next = head;
+  let length = getLength(head);
+  let curr = dummyNode;
+
+  for (let i = 1; i < length - n + 1; i++) {
+    curr = curr.next;
+  }
+  curr.next = curr.next.next;
+
+  return dummyNode.next;
+
+  // 栈
+  // return stackFn(head);
+
+  // 双指针
+  // return twoPointers(head);
+};
+
+// 计算链表长度
+function getLength(head) {
+  let curr = head;
+  let length = 0;
+  while (curr) {
+    length++;
+    curr = curr.next;
+  }
+  return length;
+}
+
+// 栈
+function stackFn(head) {
+  let dummyNode = new ListNode(-1);
+  dummyNode.next = head;
+  let curr = dummyNode;
+  let stack = [];
+  // 入栈
+  while (curr) {
+    stack.push(curr);
+    curr = curr.next;
+  }
+  // 出栈
+  for (let i = 0; i < n; i++) {
+    stack.pop();
+  }
+  // 待删除节点的前置节点
+  const prev = stack.at(-1);
+  prev.next = prev.next.next;
+
+  return dummyNode.next;
+}
+
+// 双指针
+var twoPointers = function(head, n) {
+  let dummyNode = new ListNode(-1);
+  dummyNode.next = head;
+
+  let fast = head;
+  // 指向哑节点
+  let slow = dummyNode;
+
+  // 前进 n 步
+  for (let i = 0; i < n; i++) {
+      fast = fast.next;
+  }
+  while (fast) {
+      fast = fast.next;
+      slow = slow.next;
+  }
+  // 此时，slow 指向待删除节点的前驱节点
+  slow.next = slow.next.next;
+
+  return dummyNode.next;
+};