[leetcode-32-Longest Valid Parentheses]

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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.

思路

The workflow of the solution is as below.

1. Scan the string from beginning to end.
2. If current character is ‘(‘,
   push its index to the stack. If current character is ‘)‘ and the
   character at the index of the top of stack is ‘(‘, we just find a
   matching pair so pop from the stack. Otherwise, we push the index of
   ‘)‘ to the stack.
3. After the scan is done, the stack will only
   contain the indices of characters which cannot be matched. Then
   let‘s use the opposite side - substring between adjacent indices
   should be valid parentheses.
4. If the stack is empty, the whole input
   string is valid. Otherwise, we can scan the stack to get longest
   valid substring as described in step 3.

 int longestValidParentheses(string s) 
    {
      stack<int>st;
      if(s.length()<=1)return 0;
      
      for(int i=0;i<s.length();i++)
      {
    if(st.empty())st.push(i);
    else 
    {
      if(s[i] ==‘(‘)st.push(i);
      else if(s[i]==‘)‘&&s[st.top()]==‘(‘)st.pop();
      else st.push(i);      
    }
      }
      if(st.empty()) return s.length();
      
      int a =s.length(),b =0;
      int longest = 0;
      while(!st.empty())
      {
    b=st.top();
    st.pop();
    longest = max(longest,a-b-1);
    a = b;
      }
      longest = max(longest,a);
      return longest;
    }

参考：

https://discuss.leetcode.com/topic/2289/my-o-n-solution-using-a-stack

[leetcode-32-Longest Valid Parentheses]

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原文地址：http://www.cnblogs.com/hellowooorld/p/7044761.html