20. Valid Parentheses(用栈实现括号匹配)

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

方法一：hashmap+栈

class Solution {
    public static boolean isValid(String s) {
        HashMap<Character,Character> map=new HashMap<Character,Character>();
        Stack<Character> stack=new Stack<Character>();
        map.put(‘)‘,‘(‘);
        map.put(‘]‘,‘[‘);
        map.put(‘}‘,‘{‘);
        int N=s.length();
        char [] nums=s.toCharArray();
        for(int i=0;i<N;i++){
            if(!stack.isEmpty() && map.get(nums[i])==stack.peek()){
                stack.pop();
            }else{
                stack.push(nums[i]);
            }

        }
        return stack.isEmpty() ? true :false;
    }
}

方法二：栈

class Solution {
    public static boolean isValid(String s) {
        Stack<Character> stack=new Stack<Character>();
        int N=s.length();
        char [] nums=s.toCharArray();
        for(int i=0;i<N;i++){
            if(nums[i]==‘(‘||nums[i]==‘[‘||nums[i]==‘{‘){
                stack.push(nums[i]);
            }else{
                if(stack.isEmpty()){
                    return false;
                }

                int cr=stack.pop();
                boolean a= cr==‘(‘ && nums[i]!=‘)‘;
                boolean b= cr==‘[‘ && nums[i]!=‘]‘;
                boolean c= cr==‘{‘ && nums[i]!=‘}‘;
                if(a||b||c){
                    return false;
                }
            }

        }
        return stack.isEmpty() ;
    }
}

原文地址：https://www.cnblogs.com/shaer/p/10859386.html