9月又是换工作的最佳时机。我幻想着只要换一份工作，就可以离开这个“破碎的地方”，赚更多的钱，做最舒服的事情，但事与愿违。
最近，一名女学生正在换工作。面试前她准备了很多问题。我以为她很有信心，结果却在算法上吃了大亏。
什么样的算法题能让面试官对一个女孩说出这么狠的话：你工作了3年了，这道算法题你都解不出来？
这是LeetCode上的一道算法题，旨在考察考生对“栈”数据结构的熟悉程度。我们来看一下。
给定一个仅包含字符‘(‘、‘)’、‘{‘、‘}’、‘[‘和‘]’的字符串 s，确定输入字符串是否有效。
如果满足以下条件，输入字符串有效：开括号必须由相同类型的括号括起来。左括号必须按正确的顺序关闭。
示例1：
Input: s = "()"Output: true示例2：
Input: s = "()[]{}"Output: true示例3：
Input: s = "(]"Output: false示例4：
Input: s = "([)]"Output: false实施例5：
Input: s = "{[]}"Output: true限制条件：
问题信息
如果我们真的没学过算法，也不知道那么多套路，那么通过问题和例子来获取尽可能多的信息是非常重要的。
那么，我们可以得到以下信息：
得到以上信息后，我想既然[]、{}、()是成对出现的，那我是不是可以一一消除呢？如果最后的结果是空字符串，那不是就说明符合题意了吗？
例如：
Input: s = "{[()]}"Step 1: The pair of () can be eliminated, and the result s is left with {[]}Step 2: The pair of [] can be eliminated, and the result s is left with {}Step 3: The pair of {} can be eliminated, and the result s is left with '', so it returns true in line with the meaning of the question代码：
const isValid = (s) => { while (true) { let len = s.length // Replace the string with '' one by one according to the matching pair s = s.replace('{}', '').replace('[]', '').replace('()', '') // There are two cases where s.length will be equal to len // 1. s is matched and becomes an empty string // 2. s cannot continue to match, so its length is the same as the len at the beginning, for example ({], len is 3 at the beginning, and it is still 3 after matching, indicating that there is no need to continue matching, and the result is false if (s.length === len) { return len === 0 } }}暴力消除方式还是可以通过LeetCode的用例，但是性能差了一点，哈哈。
主题信息中的第二项强调对称性。栈（后进先出）和（推入和弹出）正好相反，形成明显的对称性。
例如
Input: abcOutput: cba“abc”和“cba”是对称的，所以我们可以尝试从堆栈的角度来解析：
Input: s = "{[()]}"Step 1: read ch = {, which belongs to the left bracket, and put it into the stack. At this time, there is { in the stack.Step 2: Read ch = [, which belongs to the left parenthesis, and push it into the stack. At this time, there are {[ in the stack.Step 3: read ch = (, which belongs to the left parenthesis, and push it into the stack. At this time, there are {[( in the stack.Step 4: Read ch = ), which belongs to the right parenthesis, try to read the top element of the stack (and ) just match, and pop ( out of the stack, at this time there are {[.Step 5: Read ch = ], which belongs to the right parenthesis, try to read the top element of the stack [and ] just match, pop the [ out of the stack, at this time there are {.Step 6: Read ch = }, which belongs to the right parenthesis, try to read the top element of the stack { and } exactly match, pop { out of the stack, at this time there is still '' in the stack.Step 7: There is only '' left in the stack, s = "{[()]}" conforms to the valid bracket definition and returns true.代码
const isValid = (s) => { // The empty string character is valid if (!s) { return true } const leftToRight = { '(': ')', '[': ']', '{': '}' } const stack = [] for (let i = 0, len = s.length; i < len; i++) { const ch = s[i] // Left parenthesis if (leftToRight[ch]) { stack.push(ch) } else { // start matching closing parenthesis // 1. If there is no left parenthesis in the stack, directly false // 2. There is data but the top element of the stack is not the current closing parenthesis if (!stack.length || leftToRight[ stack.pop() ] !== ch) { return false } } } // Finally check if the stack is empty return !stack.length}虽然暴力方案符合我们的常规思维，但是堆栈结构方案会更加高效。
在面试中，算法是否应该成为评价候选人的重要指标，我们不会抱怨，但近年来，几乎每家公司都将算法纳入了前端面试中。为了拿到自己喜欢的offer，复习数据结构、刷题还是有必要的。
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