Give a string s, count the number of non-empty (contiguous) substrings that have the same number of 0‘s and 1‘s, and all the 0‘s and all the 1‘s in these substrings are grouped consecutively.
Substrings that occur multiple times are counted the number of times they occur.
Example 1:
Input: "00110011" Output: 6
Input: "10101" Output: 4
计算具有相同数量的0和1的子字符串的数量，0,1要连续
C++（42ms）：
1 class Solution { 2 public: 3 int countBinarySubstrings(string s) { 4 vector<int> vec ; 5 int res = 0 ; 6 int count = 1 ; 7 for(int i = 1 ; i <= s.size() ; i++){ 8 if (s[i] == s[i-1]){ 9 count++ ; 10 }else{ 11 vec.push_back(count) ; 12 count = 1 ; 13 } 14 } 15 for(int i = 1 ; i < vec.size() ; i++){ 16 res += min(vec[i],vec[i-1]) ; 17 } 18 return res ; 19 } 20 };
原文：http://www.cnblogs.com/mengchunchen/p/7722135.html