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.
很多人会说这道题用动规,可是用动规每次匹配后还要向前到上一个匹配跟这个匹配是否连接,时间复杂度为O(n^2),其实可以换个想法,用一个bool数组来标记已经匹配过的字符,找到最长的连续标记的长度就是所求的结果。只要遍历两遍数组,时间复杂度为O(n)。
1 class Solution { 2 public: 3 int longestValidParentheses(string s) { 4 bool *a = new bool[s.length()]; 5 memset(a, false, s.length()); 6 stack<int> st; 7 for (int i = 0; i < s.length(); ++i) { 8 if (s[i] == '(') { 9 st.push(i); 10 } else if (s[i] == ')' && !st.empty()) { 11 a[i] = true; 12 a[st.top()] = true; 13 st.pop(); 14 } 15 } 16 int max_len = 0, cur_len = 0; 17 for (int i = 0; i < s.length(); ++i) { 18 if (a[i]) ++cur_len; 19 else cur_len = 0; 20 max_len = max(max_len, cur_len); 21 } 22 return max_len; 23 } 24 };