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。
常用函数
包括打印函数和一些数学函数。
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 = _ => {}
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。
// 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);
位运算的一些小技巧
判断一个整数 x 的奇偶性：x & 1 = 1 (奇数) ， x & 1 = 0 (偶数)
求一个浮点数 x 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 x 除以 2 的 n 倍：x >> n 相当于 ~~(x / pow(2, n))
判断一个数 x 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）
// 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;
}
链表
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。
/** * 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);
}
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。
let dummy = new ListNode();
// 返回
return dummy.next;
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。
/** * @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]));
参考视频：传送门
本地运行结果
list: 1->2->5->end
list: 5->end
list: end
list: 1->end
是不是很方便！
矩阵（二维数组）
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。
/**
* 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 [row, col]
*/
const getMatrixRowAndCol = (matrix) => matrix.length === 0 ? [0, 0] : [matrix.length, matrix[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[0].length;
let i = ~~(index / col);
let j = index - i * col;
return matrix[i][j];
}
/**
* 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[0].length;
let i = ~~(index / col);
let j = index - i * col;
return matrix[i][j] = value;
}
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。
/** * @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 [r1, c1] = 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([[1],[2],[3],[4]], 2, 2);
log(x)
二叉树
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。
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[index++]);
let queue = [root];
while(index < n) {
let top = queue.shift();
let v = array[index++];
top.left = v == null ? null : new TreeNode(v);
if (index < n) {
let v = array[index++];
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 = [root];
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[res.length - 1] == null) res.pop(); // 删掉结尾的 null
return res;
}
一个例子，@leetcode 110，判断一棵二叉树是不是平衡二叉树。
/** * @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])));
二分查找
参考 C++ STL 中的 lower_bound 和 upper_bound。这两个函数真的很好用的！
/** * 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[middle] < 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[middle] > target) {
len = half;
} else {
first = middle + 1;
len = len - half - 1;
}
}
return first;
}
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。
/** * @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[lower] !== target) return [-1, -1];
return [lower, upper - 1];
};
在 VS Code 中刷 LeetCode
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
然后把文件替换为下面的代码：
{
"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 [row, col]"," */","const getMatrixRowAndCol = (matrix) => matrix.length === 0 ? [0, 0] : [matrix.length, matrix[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[0].length;"," let i = ~~(index / col);"," let j = index - i * col;"," return matrix[i][j];","}","/**"," * 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[0].length;"," let i = ~~(index / col);"," let j = index - i * col;"," return matrix[i][j] = 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[index++]);"," let queue = [root];"," while(index < n) {"," let top = queue.shift();"," let v = array[index++];"," top.left = v == null ? null : new TreeNode(v);"," if (index < n) {"," let v = array[index++];"," 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 = [root];"," 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[res.length - 1] == 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[middle] < 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[middle] > target) {"," len = half;"," } else {"," first = middle + 1;"," len = len - half - 1;"," }"," }"," return first;","}",
"$1"
],
"description": "LeetCode常用代码模板"
}
}
以后每一次写题之前，键入 @lc 就会出现提示，轻松加入代码模板。
