Java C++题解leetcode856括号的分数

class Solution {
    public int scoreOfParentheses(String s) {
        Deque<Integer> sta = new ArrayDeque<>();
        sta.addLast(0);
        for (char c : s.toCharArray()) {
            if (c == '(')
                sta.addLast(0);
            else { // 结束一个括号
                int cur = sta.pollLast(); // 取出当前分数
                sta.addLast(sta.pollLast() + Math.max(cur * 2, 1)); // 更新上级括号分数
            }
        }
        return sta.peekLast();
    }
}

class Solution {
public:
    int scoreOfParentheses(string s) {
        stack<int> sta;
        sta.push(0); // 初始0用于记录结果分数
        for (auto c : s) {
            if (c == '(')
                sta.push(0);
            else { // 结束一个括号
                int cur = sta.top(); // 取出当前分数
                sta.pop();
                sta.top() += max(cur * 2, 1); // 更新上级括号分数
            }
        }
        return sta.top();
    }
};

impl Solution {
    pub fn score_of_parentheses(s: String) -> i32 {
        let mut sta = Vec::with_capacity((s.len() >> 1) + 1);
        sta.push(0); // 初始0用于记录结果分数
        for c in s.bytes() {
            if c == b'(' {
                sta.push(0);
            }
            else {
                let cur = sta.pop().unwrap();
                *sta.last_mut().unwrap() += 1.max(cur << 1);
            }
        }
        sta[0]
    }
}

class Solution {
    public int scoreOfParentheses(String s) {
        int depth = 1, res = 0;
        for (int i = 1; i < s.length(); i++) {
            depth += (s.charAt(i) == '(' ? 1 : -1);
            if (s.charAt(i - 1) == '(' && s.charAt(i) == ')') // 分数来源
                res += 1 << depth;
        }
        return res;
    }
}

class Solution {
public:
    int scoreOfParentheses(string s) {
       int depth = 1, res = 0;
        for (int i = 1; i < s.size(); i++) {
            depth += (s[i] == '(' ? 1 : -1);
            if (s[i - 1] == '(' && s[i] == ')') // 分数来源
                res += 1 << depth;
        }
        return res;
    }
};

impl Solution {
    pub fn score_of_parentheses(s: String) -> i32 {
        let (mut depth, mut res) = (1, 0);
        let ss = s.as_bytes();
        for i in 1..s.len() {
            if (ss[i] == b'(') {
                depth += 1
            }
            else {
                depth -= 1;
                if ss[i - 1] == b'(' { // 分数来源
                    res += 1 << depth;
                }
            }
        }
        res
    }
}