【LeetCode热题100（86/100）】最长有效括号

题目地址： 链接

状态转移方程：
d p [ i + 1 ] = { d p [ i − 1 ] + 2 , 若 s [ i ] = ′ ) ′ 且 s [ i − 1 ] = ′ ( ′ d p [ i ] + 2 + d p [ i − d p [ i ] − 1 ] , 若 s [ i ] = ′ ) ′ 且 s [ i − 1 ] = ′ ) ′ 且 s [ i − 1 − d p [ i ] ] = ′ ( ′ 0 , 其他情况 dp[i+1] = \begin{cases} dp[i-1] + 2, & \text{若 } s[i] = ')' \text{ 且 } s[i-1] = '(' \\ dp[i] + 2 + dp[i - dp[i] - 1], & \text{若 } s[i] = ')' \text{ 且 } s[i-1] = ')' \text{ 且 } s[i-1-dp[i]] = '(' \\ 0, & \text{其他情况} \end{cases} dp[i+1]=⎩ ⎨ ⎧dp[i−1]+2,dp[i]+2+dp[i−dp[i]−1],0,若 s[i]=′)′ 且 s[i−1]=′(′若 s[i]=′)′ 且 s[i−1]=′)′ 且 s[i−1−dp[i]]=′(′其他情况

/*
 * @lc app=leetcode.cn id=32 lang=typescript
 *
 * [32] 最长有效括号
 */

// @lc code=start
function longestValidParentheses(s: string): number {
    let n = s.length;
    const dp = new Array(n + 1).fill(0);
    let ans = 0;
    for(let i = 1; i < n; i ++) {
        if(s[i] == ')') {
            if(s[i - 1] == '(') dp[i + 1] = dp[i - 1] + 2;
            if(s[i - 1] == ')' && s[i - 1 - dp[i]] == '(') {
                dp[i + 1] = dp[i] + 2 + dp[i - dp[i] - 1]
            }
        }
        ans = Math.max(dp[i + 1], ans);
    }
    return ans;
};
// @lc code=end