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Update 32. Longest Valid Parentheses.md

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@ -12,7 +12,9 @@
* 如果遇到左括号或者栈空，则将当前下标i压入栈； * 如果遇到左括号或者栈空，则将当前下标i压入栈；
* 如果遇到右括号且栈顶为左括号，则遇到了匹配的括号对，将栈顶弹出即可； * 如果遇到右括号且栈顶为左括号，则遇到了匹配的括号对，将栈顶弹出即可；
然后就可以计算以字符`s[i]`作为结尾的最大长度：`i - stk.top()`。 然后就可以计算以字符`s[i]`作为结尾的最大长度：
* 若栈不空，则长度为`i - stk.top()`；
* 否则，即栈空，说明前面的都能匹配，则长度为`i+1`。
时空复杂度均为O(n) 时空复杂度均为O(n)