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。
本文链接:http://www.28at.com/showinfo-26-10891-0.html面试官:你工作了3年了,这道算法题你都答不出来?
声明:本网页内容旨在传播知识,若有侵权等问题请及时与本网联系,我们将在第一时间删除处理。邮件:2376512515@qq.com
上一篇: 一文读懂分布式追踪:过去、现在和未来
下一篇: CSS实现十个功能强大的一行布局技巧