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？
有效括号
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：
- 1 <= s.length <= 104
- s 仅由括号‘()[]{}’组成
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：
- 字符串 s 的长度必须是偶数，不能是奇数（成对匹配）。
- 右括号前面必须有左括号。
方法一：暴力消除法
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。
方法二：使用“栈”来解决
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: abc
Output: 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。
最后
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。
