Valid Parentheses

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

The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.

栈法

复杂度

时间 O(N) 空间 O(N)

思路

栈最典型的应用就是验证配对情况,作为有效的括号,有一个右括号就必定有一个左括号在前面,所以我们可以将左括号都push进栈中,遇到右括号的时候再pop来消掉。这里不用担心连续不同种类左括号的问题,因为有效的括号对最终还是会有紧邻的括号对。如栈中是({[,来一个]变成({,再来一个},变成(。

注意

  • 栈在peek或者pop操作之前要验证非空,否则会抛出StackEmptyException。
  • 代码最后要判断栈的大小,如果循环结束后栈内还有元素,说明也是无效的

代码

public class Solution {
    public boolean isValid(String s) {
        Map<Character, Character> map = new HashMap<Character, Character>();
        map.put('(',')');
        map.put('[',']');
        map.put('{','}');
        Stack<Character> stk = new Stack<Character>();
        for(int i = 0; i < s.length(); i++){
            Character c = s.charAt(i);
            switch(c){
                case '(': case '[': case '{':
                    stk.push(c);
                    break;
                case ')': case ']': case '}':
                    if(stk.isEmpty() || c != map.get(stk.pop())){
                        return false;
                    }
            }
        }
        return stk.isEmpty();
    }
}

2018/2

class Solution:
    def isValid(self, s):
        """
        :type s: str
        :rtype: bool
        """
        stack = []
        for c in s:
            if c == '(' or c == '{' or c == '[':
                stack.append(c)
            elif not stack:
                return False
            elif c == ')' and stack.pop() != '(':
                return False
            elif c == '}' and stack.pop() != '{':
                return False
            elif c == ']' and stack.pop() != '[':
                return False
        return not stack

ethannnli
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