用JavaScript刷LeetCode的正确姿势

香芋發表於 2019-6-28 08:48:00

虽然很多人都觉得前端算法弱，但其实 JavaScript 也可以刷题啊！最近两个月断断续续刷完了 leetcode 前 200 的 middle + hard ，总结了一些刷题常用的模板代码。走过路过发现 bug 请指出，拯救一个辣鸡（但很帅）的少年就靠您啦！
<h2>常用函数</h2>
包括打印函数和一些数学函数。
<div class="cnblogs_code">
<pre>const _max = Math.max.bind(Math);
const _min = Math.min.bind(Math);
const _pow = Math.pow.bind(Math);
const _floor = Math.floor.bind(Math);
const _round = Math.round.bind(Math);
const _ceil = Math.ceil.bind(Math);
const log = console.log.bind(console);
//const log = _ => {}</pre>
</div>
 log 在提交的代码中当然是用不到的，不过在调试时十分有用。但是当代码里面加了很多 log 的时候，提交时还需要一个个注释掉就相当麻烦了，只要将 log 赋值为空函数就可以了。
举一个简单的例子，下面的代码是可以直接提交的。
<div class="cnblogs_code">
<pre>// 计算 1+2+...+n
// const log = console.log.bind(console);
const log = _ => {}

function sumOneToN(n) {
let sum = 0;
for (let i = 1; i <= n; i++) {
 sum += i;
 log(`i=${i}: sum=${sum}`);
}
return sum;
}

sumOneToN(10);</pre>
</div>
 
<h2 class="heading">位运算的一些小技巧</h2>
判断一个整数 <code>x</code> 的奇偶性： x & 1 = 1 <code> (奇数) ，  x & 1 = 0  (偶数)</code>
求一个浮点数 <code>x</code> 的整数部分： ~~x ，对于正数相当于  floor(x)  对于负数相当于  ceil(-x) 
计算  2 ^ n ： 1 << n  相当于  pow(2, n) 
计算一个数 <code>x</code> 除以 2 的 n 倍： x >> n  相当于  ~~(x / pow(2, n)) 
判断一个数 <code>x</code> 是 2 的整数幂（即  x = 2 ^ n ）:  x & (x - 1) = 0 
※注意※：上面的位运算只对32位带符号的整数有效，如果使用的话，一定要注意数！据！范！围！
记住这些技巧的作用：
提升运行速度 ❌
提升逼格 ✅
举一个实用的例子，快速幂（原理自行google）
<div class="cnblogs_code">
<pre>// 计算x^n n为整数
function qPow(x, n) {
let result = 1;
while (n) {
 if (n & 1) result *= x; // 同 if(n%2)
 x = x * x;
 n >>= 1; // 同 n=floor(n/2)
 }
return result;
}</pre>
</div>
 
<h2 class="heading">链表</h2>
刚开始做 LeetCode 的题就遇到了很多链表的题。恶心心。最麻烦的不是写题，是调试啊！！于是总结了一些链表的辅助函数。
<div class="cnblogs_code">
<pre>/**
* 链表节点
* @param {*} val
* @param {ListNode} next
*/
function ListNode(val, next = null) {
this.val = val;
this.next = next;
}
/**
* 将一个数组转为链表
* @param {array} a
* @return {ListNode}
*/
const getListFromArray = (a) => {
let dummy = new ListNode()
let pre = dummy;
a.forEach(x => pre = pre.next = new ListNode(x));
return dummy.next;
}
/**
* 将一个链表转为数组
* @param {ListNode} node
* @return {array}
*/
const getArrayFromList = (node) => {
let a = [];
while (node) {
 a.push(node.val);
 node = node.next;
}
return a;
}
/**
* 打印一个链表
* @param {ListNode} node
*/
const logList = (node) => {
let str = 'list: ';
while (node) {
 str += node.val + '->';
 node = node.next;
}
str += 'end';
log(str);
}</pre>
</div>
还有一个常用小技巧，每次写链表的操作，都要注意判断表头，如果创建一个空表头来进行操作会方便很多。
<div class="cnblogs_code">
<pre>let dummy = new ListNode();
// 返回
return dummy.next;</pre>
</div>
使用起来超爽哒~举个例子。@leetcode 82。题意就是删除链表中连续相同值的节点。
<div class="cnblogs_code">
<pre>/*
* @lc app=leetcode id=82 lang=javascript
*
* Remove Duplicates from Sorted List II
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var deleteDuplicates = function(head) {
// 空指针或者只有一个节点不需要处理
if (head === null || head.next === null) return head;

let dummy = new ListNode();
let oldLinkCurrent = head;
let newLinkCurrent = dummy;

while (oldLinkCurrent) {
 let next = oldLinkCurrent.next;
 // 如果当前节点和下一个节点的值相同就要一直向前直到出现不同的值
 if (next && oldLinkCurrent.val === next.val) {
 while (next && oldLinkCurrent.val === next.val) {
 next = next.next;
 }
 oldLinkCurrent = next;
 } else {
 newLinkCurrent = newLinkCurrent.next = oldLinkCurrent;
 oldLinkCurrent = oldLinkCurrent.next;
 }
}
newLinkCurrent.next = null; // 记得结尾置空~
 logList(dummy.next);
return dummy.next;
};

deleteDuplicates(getListFromArray([1,2,3,3,4,4,5]));
deleteDuplicates(getListFromArray([1,1,2,2,3,3,4,4,5]));
deleteDuplicates(getListFromArray([1,1]));
deleteDuplicates(getListFromArray([1,2,2,3,3]));</pre>
</div>
本地运行结果
<div class="cnblogs_code">
<pre>list: 1->2->5->end
list: 5->end
list: end
list: 1->end</pre>
</div>
是不是很方便！
<h2 class="heading">矩阵（二维数组）</h2>
矩阵的题目也有很多，基本每一个需要用到二维数组的题，都涉及到初始化，求行数列数，遍历的代码。于是简单提取出来几个函数。
<div class="cnblogs_code">
<pre>/**
* 初始化一个二维数组
* @param {number} r 行数
* @param {number} c 列数
* @param {*} init 初始值
*/
const initMatrix = (r, c, init = 0) => new Array(r).fill().map(_ => new Array(c).fill(init));
/**
* 获取一个二维数组的行数和列数
* @param {any[][]} matrix
* @return
*/
const getMatrixRowAndCol = (matrix) => matrix.length === 0 ? : .length];
/**
* 遍历一个二维数组
* @param {any[][]} matrix
* @param {Function} func
*/
const matrixFor = (matrix, func) => {
matrix.forEach((row, i) => {
 row.forEach((item, j) => {
 func(item, i, j, row, matrix);
 });
})
}
/**
* 获取矩阵第index个元素从0开始
* @param {any[][]} matrix
* @param {number} index
*/
function getMatrix(matrix, index) {
let col = matrix.length;
let i = ~~(index / col);
let j = index - i * col;
return matrix;
}
/**
* 设置矩阵第index个元素从0开始
* @param {any[][]} matrix
* @param {number} index
*/
function setMatrix(matrix, index, value) {
let col = matrix.length;
let i = ~~(index / col);
let j = index - i * col;
return matrix = value;
}</pre>
</div>
找一个简单的矩阵的题示范一下用法。@leetcode 566。题意就是将一个矩阵重新排列为r行c列。
<div class="cnblogs_code">
<pre>/*
* @lc app=leetcode id=566 lang=javascript
*
* Reshape the Matrix
*/
/**
* @param {number[][]} nums
* @param {number} r
* @param {number} c
* @return {number[][]}
*/
var matrixReshape = function(nums, r, c) {
// 将一个矩阵重新排列为r行c列
// 首先获取原来的行数和列数
let = getMatrixRowAndCol(nums);
log(r1, c1);
// 不合法的话就返回原矩阵
if (!r1 || r1 * c1 !== r * c) return nums;
// 初始化新矩阵
let matrix = initMatrix(r, c);
// 遍历原矩阵生成新矩阵
matrixFor(nums, (val, i, j) => {
 let index = i * c1 + j; // 计算是第几个元素
 log(index);
 setMatrix(matrix, index, val); // 在新矩阵的对应位置赋值
 });
return matrix;
};

let x = matrixReshape([,,,], 2, 2);
log(x)</pre>
</div>
 
<h2 class="heading">二叉树</h2>
当我做到二叉树相关的题目，我发现，我错怪链表了，呜呜呜这个更恶心。
当然对于二叉树，只要你掌握先序遍历，后序遍历，中序遍历，层序遍历，递归以及非递归版，先序中序求二叉树，先序后序求二叉树，基本就能AC大部分二叉树的题目了（我瞎说的）。
二叉树的题目 input 一般都是层序遍历的数组，所以写了层序遍历数组和二叉树的转换，方便调试。
<div class="cnblogs_code">
<pre>function TreeNode(val, left = null, right = null) {
this.val = val;
this.left = left;
this.right = right;
}
/**
* 通过一个层次遍历的数组生成一棵二叉树
* @param {any[]} array
* @return {TreeNode}
*/
function getTreeFromLayerOrderArray(array) {
let n = array.length;
if (!n) return null;
let index = 0;
let root = new TreeNode(array);
let queue = ;
while(index < n) {
 let top = queue.shift();
 let v = array;
 top.left = v == null ? null : new TreeNode(v);
 if (index < n) {
 let v = array;
 top.right = v == null ? null : new TreeNode(v);
 }
 if (top.left) queue.push(top.left);
 if (top.right) queue.push(top.right);
}
return root;
}
/**
* 层序遍历一棵二叉树生成一个数组
* @param {TreeNode} root
* @return {any[]}
*/
function getLayerOrderArrayFromTree(root) {
let res = [];
let que = ;
while (que.length) {
 let len = que.length;
 for (let i = 0; i < len; i++) {
 let cur = que.shift();
 if (cur) {
 res.push(cur.val);
 que.push(cur.left, cur.right);
 } else {
 res.push(null);
 }
 }
}
while (res.length > 1 && res == null) res.pop(); // 删掉结尾的 null
return res;
}</pre>
</div>
 
一个例子，@leetcode 110，判断一棵二叉树是不是平衡二叉树。
<div class="cnblogs_code">
<pre>/**
* @param {TreeNode} root
* @return {boolean}
*/
var isBalanced = function(root) {
if (!root) return true; // 认为空指针也是平衡树吧

// 获取一个二叉树的深度
const d = (root) => {
 if (!root) return 0;
 return _max(d(root.left), d(root.right)) + 1;
}

let leftDepth = d(root.left);
let rightDepth = d(root.right);

// 深度差不超过 1 且子树都是平衡树
if (_min(leftDepth, rightDepth) + 1 >= _max(leftDepth, rightDepth)
 && isBalanced(root.left) && isBalanced(root.right)) return true;

return false;
};

log(isBalanced(getTreeFromLayerOrderArray([3,9,20,null,null,15,7])));
log(isBalanced(getTreeFromLayerOrderArray([1,2,2,3,3,null,null,4,4])));</pre>
</div>
 
<h2 class="heading">二分查找</h2>
参考 C++ STL 中的 lower_bound  和 upper_bound 。这两个函数真的很好用的！
<div class="cnblogs_code">
<pre>/**
* 寻找>=target的最小下标
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
function lower_bound(nums, target) {
let first = 0;
let len = nums.length;

while (len > 0) {
 let half = len >> 1;
 let middle = first + half;
 if (nums < target) {
 first = middle + 1;
 len = len - half - 1;
 } else {
 len = half;
 }
}
return first;
}

/**
* 寻找>target的最小下标
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
function upper_bound(nums, target) {
let first = 0;
let len = nums.length;

while (len > 0) {
 let half = len >> 1;
 let middle = first + half;
 if (nums > target) {
 len = half;
 } else {
 first = middle + 1;
 len = len - half - 1;
 }
}
return first;
}</pre>
</div>
照例，举个例子，@leetcode 34。题意是给一个排好序的数组和一个目标数字，求数组中等于目标数字的元素最小下标和最大下标。不存在就返回 -1。
<div class="cnblogs_code">
<pre>/*
* @lc app=leetcode id=34 lang=javascript
*
* Find First and Last Position of Element in Sorted Array
*/
/**
* @param {number[]} nums
* @param {number} target
* @return {number[]}
*/
var searchRange = function(nums, target) {
let lower = lower_bound(nums, target);
let upper = upper_bound(nums, target);
let size = nums.length;
// 不存在返回 [-1, -1]
if (lower >= size || nums !== target) return [-1, -1];
return ;
};</pre>
</div>
 
<h2 class="heading">在 VS Code 中刷 LeetCode</h2>
前面说的那些模板，难道每一次打开新的一道题都要复制一遍么？当然不用啦。
首先配置代码片段选择 Code -> Preferences -> User Snippets ，然后选择 JavaScript
 
<img src="https://user-gold-cdn.xitu.io/2019/6/22/16b7e9c0a793f7d9?w=541&h=254&f=png&s=128089" alt="">
 
然后把文件替换为下面的代码：
<div class="cnblogs_code">
<pre>{
"leetcode template": {
"prefix": "@lc",
"body": [
 "const _max = Math.max.bind(Math);","const _min = Math.min.bind(Math);","const _pow = Math.pow.bind(Math);","const _floor = Math.floor.bind(Math);","const _round = Math.round.bind(Math);","const _ceil = Math.ceil.bind(Math);","const log = console.log.bind(console);","// const log = _ => {}","/**************** 链表 ****************/","/**"," * 链表节点"," * @param {*} val"," * @param {ListNode} next"," */","function ListNode(val, next = null) {"," this.val = val;"," this.next = next;","}","/**"," * 将一个数组转为链表"," * @param {array} array"," * @return {ListNode}"," */","const getListFromArray = (array) => {"," let dummy = new ListNode()"," let pre = dummy;"," array.forEach(x => pre = pre.next = new ListNode(x));"," return dummy.next;","}","/**"," * 将一个链表转为数组"," * @param {ListNode} list"," * @return {array}"," */","const getArrayFromList = (list) => {"," let a = [];"," while (list) {"," a.push(list.val);"," list = list.next;"," }"," return a;","}","/**"," * 打印一个链表"," * @param {ListNode} list "," */","const logList = (list) => {"," let str = 'list: ';"," while (list) {"," str += list.val + '->';"," list = list.next;"," }"," str += 'end';"," log(str);","}","/**************** 矩阵（二维数组） ****************/","/**"," * 初始化一个二维数组"," * @param {number} r 行数"," * @param {number} c 列数"," * @param {*} init 初始值"," */","const initMatrix = (r, c, init = 0) => new Array(r).fill().map(_ => new Array(c).fill(init));","/**"," * 获取一个二维数组的行数和列数"," * @param {any[][]} matrix"," * @return "," */","const getMatrixRowAndCol = (matrix) => matrix.length === 0 ? : .length];","/**"," * 遍历一个二维数组"," * @param {any[][]} matrix "," * @param {Function} func "," */","const matrixFor = (matrix, func) => {"," matrix.forEach((row, i) => {"," row.forEach((item, j) => {"," func(item, i, j, row, matrix);"," });"," })","}","/**"," * 获取矩阵第index个元素从0开始"," * @param {any[][]} matrix "," * @param {number} index "," */","function getMatrix(matrix, index) {"," let col = matrix.length;"," let i = ~~(index / col);"," let j = index - i * col;"," return matrix;","}","/**"," * 设置矩阵第index个元素从0开始"," * @param {any[][]} matrix "," * @param {number} index "," */","function setMatrix(matrix, index, value) {"," let col = matrix.length;"," let i = ~~(index / col);"," let j = index - i * col;"," return matrix = value;","}","/**************** 二叉树 ****************/","/**"," * 二叉树节点"," * @param {*} val"," * @param {TreeNode} left"," * @param {TreeNode} right"," */","function TreeNode(val, left = null, right = null) {"," this.val = val;"," this.left = left;"," this.right = right;","}","/**"," * 通过一个层次遍历的数组生成一棵二叉树"," * @param {any[]} array"," * @return {TreeNode}"," */","function getTreeFromLayerOrderArray(array) {"," let n = array.length;"," if (!n) return null;"," let index = 0;"," let root = new TreeNode(array);"," let queue = ;"," while(index < n) {"," let top = queue.shift();"," let v = array;"," top.left = v == null ? null : new TreeNode(v);"," if (index < n) {"," let v = array;"," top.right = v == null ? null : new TreeNode(v);"," }"," if (top.left) queue.push(top.left);"," if (top.right) queue.push(top.right);"," }"," return root;","}","/**"," * 层序遍历一棵二叉树生成一个数组"," * @param {TreeNode} root "," * @return {any[]}"," */","function getLayerOrderArrayFromTree(root) {"," let res = [];"," let que = ;"," while (que.length) {"," let len = que.length;"," for (let i = 0; i < len; i++) {"," let cur = que.shift();"," if (cur) {"," res.push(cur.val);"," que.push(cur.left, cur.right);"," } else {"," res.push(null);"," }"," }"," }"," while (res.length > 1 && res == null) res.pop(); // 删掉结尾的 null"," return res;","}","/**************** 二分查找 ****************/","/**"," * 寻找>=target的最小下标"," * @param {number[]} nums"," * @param {number} target"," * @return {number}"," */","function lower_bound(nums, target) {"," let first = 0;"," let len = nums.length;",""," while (len > 0) {"," let half = len >> 1;"," let middle = first + half;"," if (nums < target) {"," first = middle + 1;"," len = len - half - 1;"," } else {"," len = half;"," }"," }"," return first;","}","","/**"," * 寻找>target的最小下标"," * @param {number[]} nums"," * @param {number} target"," * @return {number}"," */","function upper_bound(nums, target) {"," let first = 0;"," let len = nums.length;",""," while (len > 0) {"," let half = len >> 1;"," let middle = first + half;"," if (nums > target) {"," len = half;"," } else {"," first = middle + 1;"," len = len - half - 1;"," }"," }"," return first;","}",
 "$1"
],
"description": "LeetCode常用代码模板"
}
}</pre>
</div>
 
以后每一次写题之前，键入 <code>@lc</code> 就会出现提示，轻松加入代码模板。 
<img src="https://user-gold-cdn.xitu.io/2019/6/22/16b7ea5270447a30?w=508&h=111&f=png&s=15530" alt="">
 
当然，必须推荐刷题神器，vscode 中的一款插件 vscode-leetcode
最后我要大声说，前端真的有机会用到算法的（不只面试）！来一起快乐刷题！ 
来源：https://www.cnblogs.com/wenruo/p/11100537.html

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用JavaScript刷LeetCode的正确姿势