update 4 leetcode

Author: kyriejoshua

leetcode/141.linked-list-cycle.js

@@ -0,0 +1,59 @@
+/*
+ * @lc app=leetcode id=141 lang=javascript
+ *
+ * [141] Linked List Cycle
+ */
+
+// @lc code=start
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ * this.val = val;
+ * this.next = null;
+ * }
+ */
+
+/**
+ * 快慢指针
+ * @param {ListNode} head
+ * @return {boolean}
+ */
+var hasCycle = function(head) {
+ if (head === null || head.next === null) {
+ return false;
+ }
+ let slow = fast = head;
+ fast = fast.next;
+
+ // 如果快指针的下一个节点和慢指针重合，则说明形成了环
+ while (slow !== null) {
+ if (fast === null || fast.next === null) {
+ return false;
+ }
+ if (slow === fast.next) {
+ return true;
+ }
+
+ slow = slow.next;
+ fast = fast.next.next;
+ }
+};
+
+/**
+ * 哈希表的方式 O(n)
+ * @param {*} head
+ * @returns
+ */
+var hasCycle2 = function(head) {
+ let map = new Map();
+
+ while (head !== null) {
+ if (map.get(head.val) === head.next) {
+ return true;
+ }
+ map.set(head.val, head.next);
+ head = head.next;
+ }
+ return false;
+};
+// @lc code=end

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

@@ -0,0 +1,45 @@
+/*
+ * @lc app=leetcode id=19 lang=javascript
+ *
+ * [19] Remove Nth Node From End of List
+ */
+
+// @lc code=start
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ * this.val = (val===undefined ? 0 : val)
+ * this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * 可以遍历一次获取索引，再遍历得到倒数索引
+ * 也可以使用双指针，快慢指针的方式
+ * @param {ListNode} head
+ * @param {number} n
+ * @return {ListNode}
+ */
+var removeNthFromEnd = function(head, n) {
+ let first = head;
+ let prev = new ListNode(0, head);
+ let second = prev;
+
+ // 第一个链表先遍历到 n 的位置，再往后遍历，就正好是 length - n 的数据
+ for (let i = 0; i < n; i++) {
+ first = first.next;
+ }
+
+ // 第一个链表到结尾的时候，第二个正好到倒数 n 的位置
+ while (first !== null) {
+ first = first.next;
+ second = second.next;
+ }
+
+ // 删除节点
+ second.next = second.next.next;
+
+ // 返回删除之后的链表
+ const ans = prev.next;
+ return ans;
+};
+// @lc code=end

leetcode/206.reverse-linked-list.js

@@ -0,0 +1,37 @@
+/*
+ * @lc app=leetcode id=206 lang=javascript
+ *
+ * [206] Reverse Linked List
+ */
+
+// @lc code=start
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val, next) {
+ * this.val = (val===undefined ? 0 : val)
+ * this.next = (next===undefined ? null : next)
+ * }
+ */
+/**
+ * 迭代的方式
+ * @param {ListNode} head
+ * @return {ListNode}
+ */
+var reverseList = function(head) {
+ let curr = head;
+ let prev = null;
+
+ while (curr !== null) {
+ // 保存下一节点，后面这个指针会改变
+ const next = curr.next;
+ // 反过来把下一节点保存到前一个
+ curr.next = prev;
+ // 把当前节点当做是前一个几点
+ prev = curr;
+ // 往后迭代原来的节点
+ curr = next;
+ }
+
+ return prev;
+};
+// @lc code=end

leetcode/697.degree-of-an-array.js

@@ -0,0 +1,75 @@
+/*
+ * @lc app=leetcode id=697 lang=javascript
+ *
+ * [697] Degree of an Array
+ */
+
+// @lc code=start
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var findShortestSubArray = function(nums) {
+ // 遍历保存所有数的数量及索引
+ let map = new Map();
+ // 保存所有数的数量，数量和相应的出现索引
+ for (let i = 0; i < nums.length; i++) {
+ const ele = nums[i];
+ let current = map.get(ele) || {};
+ current.count = (current.count || 0) + 1;
+ current.firstIndex = current.firstIndex !== undefined ? current.firstIndex : i;
+ current.lastIndex = i;
+ map.set(ele, current);
+ }
+
+ // 保存数组的度，和对应度的数组长度，比下面节约使用了数组
+ let maxCount = 0;
+ let minLength = Number.MAX_SAFE_INTEGER;
+ // 注意可能会有多个数组的度
+ for (const [key, obj] of map) {
+ if (obj.count > maxCount) {
+ // 数组的长度是差值加 1
+ minLength = obj.lastIndex - obj.firstIndex + 1;
+ maxCount = obj.count;
+ }
+ if (obj.count === maxCount) {
+ minLength = (obj.lastIndex - obj.firstIndex + 1) < minLength ? (obj.lastIndex - obj.firstIndex + 1) : minLength;
+ }
+ }
+
+ return minLength;
+};
+
+/**
+ * 数组的方式保存
+ * @param {*} nums
+ * @returns
+ */
+var findShortestSubArray2 = function(nums) {
+ // 遍历保存所有数的数量及索引
+ let map = new Map();
+ // 保存所有数的数量，数量和相应的出现索引
+ for (let i = 0; i < nums.length; i++) {
+ const ele = nums[i];
+ let current = map.get(ele) || {};
+ current.count = (current.count || 0) + 1;
+ current.firstIndex = current.firstIndex !== undefined ? current.firstIndex : i;
+ current.lastIndex = i;
+ map.set(ele, current);
+ }
+
+ let maxEle = [{ count: 0 }];
+ // 保存数组的度的索引，注意可能会有多个数组的度，所以需要数组来保存
+ for (const [key, obj] of map) {
+ if (obj.count > maxEle[0].count) {
+ maxEle = [obj];
+ } else if (obj.count === maxEle[0].count) {
+ maxEle.push(obj);
+ }
+ }
+
+ return maxEle.reduce((prev, cur) => {
+ return (cur.lastIndex - cur.firstIndex + 1) < prev ? (cur.lastIndex - cur.firstIndex + 1) : prev;
+ }, Number.MAX_SAFE_INTEGER);
+};
+// @lc code=end