Valid Parentheses (E)

题目

Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine if the input string is valid.

An input string is valid if:

1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.

Note that an empty string is also considered valid.

Example 1:

Input: "()"
Output: true

Example 2:

Input: "()[]{}"
Output: true

Example 3:

Input: "(]"
Output: false

Example 4:

Input: "([)]"
Output: false

Example 5:

Input: "([)]"
Output: false

题意

给定一个只包含括号的字符串，判断括号是否匹配。

思路

用栈处理：是左半边括号则压入栈，是右半边括号则比较与栈顶是否匹配。最后检查栈是否清空。

代码实现

Java

class Solution {
    public boolean isValid(String s) {
        Deque<Character> stack = new ArrayDeque<>();
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == ‘(‘ || c == ‘[‘ || c == ‘{‘) {
                stack.push(c);
            } else if (c == ‘)‘) {
                if (!stack.isEmpty() && stack.peek() == ‘(‘) {
                    stack.pop();
                } else {
                    return false;
                }
            } else if (c == ‘]‘) {
                if (!stack.isEmpty() && stack.peek() == ‘[‘) {
                    stack.pop();
                } else {
                    return false;
                }
            } else {
                if (!stack.isEmpty() && stack.peek() == ‘{‘) {
                    stack.pop();
                } else {
                    return false;
                }
                    
            }
        }
        return stack.isEmpty();
    }
}

JavaScript

/**
 * @param {string} s
 * @return {boolean}
 */
var isValid = function (s) {
  let stack = []

  for (let c of s.split(‘‘)) {
    switch (c) {
      case ‘(‘:
      case ‘[‘:
      case ‘{‘:
        stack.push(c)
        break
      case ‘)‘:
        if (stack.length === 0 || stack.pop() !== ‘(‘) {
          return false
        }
        break
      case ‘]‘:
        if (stack.length === 0 || stack.pop() !== ‘[‘) {
          return false
        }
        break
      case ‘}‘:
        if (stack.length === 0 || stack.pop() !== ‘{‘) {
          return false
        }
        break
    }
  }

  return stack.length === 0
}

0020. Valid Parentheses (E)

原文：https://www.cnblogs.com/mapoos/p/13171278.html