Input: s = "()"Output: true
Input: s = "()[]{}"Output: true
Input: s = "(]"Output: false
Input: s = "([)]"Output: false
Input: s = "{[]}"Output: true
1 <= s.length <= 104 s 仅由括号‘()[]{}’组成
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。
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
const isValid = (s) => {while (true) {let len = s.length// Replace the string with '' one by one according to the matching pairs = 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 falseif (s.length === len) {return len === 0}}}
Input: abcOutput: cba
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.
const isValid = (s) => {// The empty string character is validif (!s) {return true}const leftToRight = {'(': ')','[': ']','{': '}'}const stack = []for (let i = 0, len = s.length; i < len; i++) {const ch = s[i]// Left parenthesisif (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 parenthesisif (!stack.length || leftToRight[ stack.pop() ] !== ch) {return false}}}// Finally check if the stack is emptyreturn !stack.length}
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