leetcode刷题（30）——20. 有效的括号

class Solution {
    public boolean isValid(String s) {
        int left1=0;
        int left2=0;
        int left3=0;
        int right1=0;
        int right2=0;
        int right3=0;
        for(int i=0;i<s.length();i++){
            char c = s.charAt(i);
            if(c=='['){
                left1++;
            }else if(c==']'){
                right1++;
            }else if(c=='('){
                left2++;
            }else if(c==')'){
                right2++;
            }else if(c=='{'){
                left3++;
            }else if(c=='}'){
                right3++;
            } 
        }
        return left1==right1&&left2==right2&&left3==right3;

    }
}

private HashMap<Character, Character> mappings;

        // Initialize hash map with mappings. This simply makes the code easier to read.
        public Solution() {
            this.mappings = new HashMap<Character, Character>();
            this.mappings.put(')', '(');
            this.mappings.put('}', '{');
            this.mappings.put(']', '[');
        }

        public boolean isValid(String s) {

            // Initialize a stack to be used in the algorithm.
            Stack<Character> stack = new Stack<Character>();

            for (int i = 0; i < s.length(); i++) {
                char c = s.charAt(i);

                // If the current character is a closing bracket.
                if (this.mappings.containsKey(c)) {

                    // Get the top element of the stack. If the stack is empty, set a dummy value of '#'
                    char topElement = stack.empty() ? '#' : stack.pop();

                    // If the mapping for this bracket doesn't match the stack's top element, return false.
                    if (topElement != this.mappings.get(c)) {
                        return false;
                    }
                } else {
                    // If it was an opening bracket, push to the stack.
                    stack.push(c);
                }
            }

            // If the stack still contains elements, then it is an invalid expression.
            return stack.isEmpty();
        }