leetcode 20. Valid Parentheses

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

经典的括号问题，61B课上讲过stack方法。

class Solution {
    public boolean isValid(String s) {
        if(s.length() == 0) return true;
        if(s.length() % 2 != 0) return false;
        Map<Character,Integer> map = new HashMap<>();
        map.put('(',1);
        map.put(')',-1);
        map.put('{',2);
        map.put('}',-2);
        map.put('[',3);
        map.put(']',-3);
        
        Stack<Character> stack = new Stack<>();
        for(int i=0; i<s.length(); i++) {
            char a = s.charAt(i);
            if(map.get(a) > 0) stack.push(a);
            if(map.get(a) < 0) {
                if(stack.empty()) return false;
                char b = stack.pop();
                if((map.get(a) + map.get(b)) != 0) return false;
            }
        }
        if(!stack.empty()) return false;
        return true;
    }
}

 

解法二：同样用stack，但是没有用哈希表，而且是当指针指向左括号时，选择将相应的右括号放入stack中

public boolean isValid(String s) {
    Stack<Character> stack = new Stack<Character>();
    for (char c : s.toCharArray()) {
        if (c == '(')
            stack.push(')');
        else if (c == '{')
            stack.push('}');
        else if (c == '[')
            stack.push(']');
        else if (stack.isEmpty() || stack.pop() != c)
            return false;
    }
    return stack.isEmpty();
}