最长有效括号

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。

方法一、栈

由于要存在配对，必然有')'存在，配对的长度为上一个')'到当前')'之间的距离，即为最长配对子串，因此，我们使用栈结构，存储上一次配对的')'所在的索引。栈存放**[最后一个没有被匹配的右括号的下标]**

复杂度分析
时间复杂度：O(n)，n为字符串的长度，遍历一次
空间复杂度：O(n)，n为字符串的长度，栈最多存储n个字符

Swift

func longestValidParentheses(_ s: String) -> Int {let cnt = s.countvar res:Int = 0//栈，存放[最后一个没有被匹配的右括号的下标]let stack = Stack<Int>()stack.push(-1)//为了统一期间，假数据填充for i in 0..<cnt {let chara = s[s.index(s.startIndex, offsetBy: i)]if chara == "(" {stack.push(i)}else if(chara == ")") {let _ = stack.pop()if stack.count() == 0 {stack.push(i)}else {res = max(res, i - (stack.top() ?? 0))}}}return res}

OC

- (NSInteger)longestValidParentheses:(NSString *)s {NSInteger res = 0;//存放[最后一个没有被匹配的右括号的下标]StackOC *stack = [[StackOC alloc] init];[stack push:@(-1)];//for (NSInteger i=0; i<s.length; i++) {unichar character = [s characterAtIndex:i];if (character == '(') {[stack push:@(i)];}else if(character == ')') {[stack pop];if ([stack empty]) {[stack push:@(i)];}else {//当前右括号索引减去上一个右括号的索引，表示最长串res = MAX(res, i-[[stack top] integerValue]);}}}return res;
}

方法二、双指针+双向遍历

思想：left对应 '('个数，right对应')'个数，当两者相等时，说明配对成功。
边界处理：当从左向右遍历时，right>left表示不连续了，要从0开始计数，因此将left和right置为0。
当从右向左遍历时，当left>right时表示不连续了，计数归0

复杂度分析
时间复杂度：O(n)，n为字符串的长度，遍历一次
空间复杂度：O(1)

Swift

//双指针法，空间复杂度为O(1)func longestValidParentheses(_ s: String) -> Int {let cnt = s.countvar left = 0, right = 0, maxLen = 0//从左向右找一次for i in 0..<cnt {let str = s[s.index(s.startIndex, offsetBy: i)]if str == "(" {left += 1}else if(str == ")") {right += 1}if left == right {maxLen = max(maxLen, 2*right)}else if(right > left) {left = 0right = 0}}//从右向左找一次left = 0; right = 0var j = cnt - 1while j > 0 {let str = s[s.index(s.startIndex, offsetBy: j)]if str == "(" {left += 1}else if(str == ")") {right += 1}if left == right {maxLen = max(maxLen, 2*right)}else if(left > right) {left = 0right = 0}j -= 1}return maxLen}

OC

//双指针+左右遍历
- (NSInteger)longestValidParentheses:(NSString *)s {NSInteger left = 0, right = 0, maxLen = 0;//从左向右匹配一次for (NSInteger i=0; i<s.length; i++) {unichar charater = [s characterAtIndex:i];if (charater == '(') {left++;}else if(charater == ')') {right++;}if (left == right) {maxLen = MAX(maxLen, 2 * right);}else if(right > left) {left = right = 0;}}//从右向左匹配一次for (NSInteger i=s.length-1; i>0; i--) {unichar charater = [s characterAtIndex:i];if (charater == '(') {left++;}else if(charater == ')') {right++;}if (left == right) {maxLen = MAX(maxLen, 2*left);}else if(left > right) {left = right = 0;}}return maxLen;
}