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: true

解法思路（一）

解法描述

• 对于待判断的字符串，逐字符的判断；
• 如果属于左括号，就压入栈中；
• 如果属于右括号，就和栈顶的左括号比较，当然，前提是栈不为空；
  • 如果栈顶左括号和待比较的右括号不匹配，那么这个字符串就是非法的；
  • 如果最后一个右括号也能和栈顶的左括号匹配，那么就得看栈中还有没有左括号剩余，没剩余，待判断的字符串才是合法的；

解法实现（一）

关键字

栈 括号匹配

代码实现

import java.util.Stack;

class Solution20_1 {

    public boolean isValid(String s) {
    
        Stack<Character> stack = new Stack<>();

        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == '(' || c == '[' || c == '{') {
                stack.push(c);
            } else {
                if (stack.isEmpty()) {
                    return false;
                }
                char top = stack.pop();
                if (c == ')' && top != '(') {
                    return false;
                }
                if (c == ']' && top != '[') {
                    return false;
                }
                if (c == '}' && top != '{') {
                    return false;
                }
            }
        }

        return stack.isEmpty();
    }

}

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