Leetcode20. Valid parentheses

Valid parenthesis  

 -->20. Valid parenthesis

This is also a classic problem of stack application

Input ：s = "( )"          Output ：true        Input ：s = "( )[ ]{ }"        Output ：true
Input ：s = "( ]"          Output ：false        Input ：s = "( [ ) ]"          Output ：false
Input ：s = "{ [ ] }"        Output ：true

There are three kinds of brackets in this question 【】{} （） Then these three can be mixed or not , See if it matches ？

Ideas ：

The left parentheses are stacked , Then check whether the traversal character matches it , If it matches, pop up the left bracket at the top of the stack and continue to traverse , If there are elements in the last stack , Or there is a string that has not been traversed, but the stack is empty, which is not matched

class Solution {
    public boolean isValid(String s) {
       // If it's a left bracket, put it on the stack 
       Stack<Character> stack = new Stack<>();
       for(int i =0;i<s.length();++i){
            // If it's a left bracket, put it on the stack 
            if(s.charAt(i)=='('||s.charAt(i)=='['
            ||s.charAt(i)=='{'){
                stack.push(s.charAt(i));
            }else {
                // If it's not the right parenthesis, compare it with the top element of the stack 
                if(stack.empty()){
                    // If the stack is empty , Can't compare 
                    return false;
                }else {
                    // The stack is not empty to see whether it is a match 
                    if((stack.peek()=='('&&s.charAt(i)==')')
                    ||(stack.peek()=='{'&&s.charAt(i)=='}')||
                    (stack.peek()=='['&&s.charAt(i)==']')){
                        stack.pop();
                    }else {
                        // If equal, pop up the top element 
                       return false;
                    }
                }
            }
       }
           return stack.empty();
    }
}