题目
给你一个只包含 '(' 和 ')' 的字符串,找出最长有效(格式正确且连续)括号子串的长度。
约束
解法
解法一
class Solution {
public:
int longestValidParentheses(string s) {
int maxans = 0, n = s.length();
vector<int> dp(n, 0);
for (int i = 1; i < n; i++) {
if (s[i] == ')') {
if (s[i - 1] == '(') {
dp[i] = (i >= 2 ? dp[i - 2] : 0) + 2;
} else if (i - dp[i - 1] > 0 && s[i - dp[i - 1] - 1] == '(') {
dp[i] = dp[i - 1] + ((i - dp[i - 1]) >= 2 ? dp[i - dp[i - 1] - 2] : 0) + 2;
}
maxans = max(maxans, dp[i]);
}
}
return maxans;
}
};
解法二
class Solution {
public:
int longestValidParentheses(string s) {
int maxans = 0;
stack<int> stk;
stk.push(-1);
for (int i = 0; i < s.length(); i++) {
if (s[i] == '(') {
stk.push(i);
} else {
stk.pop();
if (stk.empty()) {
stk.push(i);
} else {
maxans = max(maxans, i - stk.top());
}
}
}
return maxans;
}
};
解法三
class Solution {
public:
int longestValidParentheses(string s) {
int left = 0, right = 0, maxlength = 0;
for (int i = 0; i < s.length(); i++) {
if (s[i] == '(') {
left++;
} else {
right++;
}
if (left == right) {
maxlength = max(maxlength, 2 * right);
} else if (right > left) {
left = right = 0;
}
}
left = right = 0;
for (int i = (int)s.length() - 1; i >= 0; i--) {
if (s[i] == '(') {
left++;
} else {
right++;
}
if (left == right) {
maxlength = max(maxlength, 2 * left);
} else if (left > right) {
left = right = 0;
}
}
return maxlength;
}
};
标签:right,int,maxans,最长,括号,stk,LeetCode,dp,left
From: https://www.cnblogs.com/chuixulvcao/p/17094008.html