括号串合法性判断（栈实现）

class Solution {
    public boolean isValid(String s) {
        boolean result = true;
        char[] charsArray = s.toCharArray();
        int stringLength = charsArray.length;
//        长度小于2或者长度不为偶数判定为假
        if (stringLength < 2 || (stringLength & 1) == 1) {
            result = false;
            return result;
        }
        Stack<Character> characterStack = new Stack<>();
        HashSet<Character> leftSet = new HashSet<>();
        HashSet<Character> rightSet = new HashSet<>();
        HashMap pairHashMap = new HashMap<Character, Character>();
        leftSet.add('(');
        leftSet.add('{');
        leftSet.add('[');
        rightSet.add(')');
        rightSet.add('}');
        rightSet.add(']');
        pairHashMap.put(')', '(');
        pairHashMap.put('}', '{');
        pairHashMap.put(']', '[');
//        遍历字符串数组指针
        int pointA = 0;
//        开始遍历字符串数组
        while (pointA != stringLength) {
//            当前指针所指的字符串数组元素
            char nowItem = charsArray[pointA];
//            对nowItem情况三种进行处理：1：开括号、闭括号、其他字符
            if (leftSet.contains(nowItem)) {
                characterStack.push(nowItem);
            } else if (!characterStack.isEmpty() && rightSet.contains(nowItem)) {
                Character topItem = characterStack.pop();
                if (!(pairHashMap.get(nowItem) == topItem))
                    return false;
            } else
                return false;
            pointA++;
        }
//        如果栈中还剩下元素，则是有开括号未找到匹配
        if (!characterStack.isEmpty())
            result = false;
        return result;
    }
}

执行用时：2 ms, 在所有 Java 提交中击败了74.39% 的用户
内存消耗：36.6 MB, 在所有 Java 提交中击败了46.31% 的用户

标签：合法性,characterStack,nowItem,rightSet,括号,add,判断,result,leftSet
来源： https://blog.csdn.net/qq_40131916/article/details/119253141