【算法】【Dynamic Programming】Longest Valid Parentheses

Difficulty: Hard

Description

Given a string containing just the characters ‘(’ and ‘)’, find the length of the longest valid (well-formed) parentheses substring.

For “(()”, the longest valid parentheses substring is “()”, which has length = 2.

Another example is “)()())”, where the longest valid parentheses substring is “()()”, which has length = 4.

Solution

思路

自己的实现

1. 找到(开头的字符，一直往下扫描；
2. 若(的数量多于或等于)的数量，继续扫描（字符串直到目前合法）；
3. 若(的数量少于)的数量，说明不合法，记录之前合法字符串最大长度，回到第1步。

自己的实现有个缺点：扫描到末尾，要从末尾往回再扫描一部分，因为要处理类似(()的情况。

class Solution {public:    int longestValidParentheses(string s) {        int len = s.length();        int i = 0, j = -1, max = 0, left = 1, right = 0;        for (; i < len && j < len;) {            i = j + 1;            // 找到'('开头            for (; i < len && s[i] == ')'; ++i) {}            if (i < len) {                for (j = i + 1; j < len; ++j) {                    if (s[j] == '(') ++left;                    else ++right;                    if (left < right) {                        max = (max < j - i) ? j - i : max;                        left = 1;                        right = 0;                        break;                    }                }            }        }        // 处理最后的一串括号        --j;        while (j > i) {            for (; j > i && s[j] == '('; --j) {}            if (j > i) {                for (right = 0, left = 0; j >= i; --j) {                    if (s[j] == ')') right++;                    else left++;                    if (left > right) break;                }            }            max = (max < 2 * right) ? 2 * right : max;        }        return max;    }};

更好的实现

使用栈，把字符所属的下标一个一个压入栈，若先后压入的下标分别代表(和)，则弹出这两个下标，目前下标减去栈顶的下标，则是目前的maxLength。

class Solution {public:    int longestValidParentheses(string s) {        stack<int>stackk;        int maxLength = 0;        //(()        stackk.push(-1);        for (int i = 0; i < s.size(); ++i) {            int t = stackk.top();            if (t != -1 && s[i] == ')' && s[t] == '(') {                stackk.pop();                maxLength = max(maxLength, i - stackk.top());            }            else {                stackk.push(i);            }        }        return maxLength;    }    int max(int a, int b) {        if (a < b) return b;        return a;    }};