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
Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01".

Notice that some of these substrings repeat and are counted the number of times they occur.

Also, "00110011" is not a valid substring because all the 0's (and 1's) are not grouped together.

Example 2:

Input: "10101"
Output: 4
Explanation: There are 4 substrings: "10", "01", "10", "01" that have equal number of consecutive 1's and 0's.

Note:

遍历s, 一开始我想到了剪枝，比如说1100，这样的话就有2个符合条件的，1100和10，但是我就把 index += 2，做了这样一个剪枝。结果虽然过了，但是非常慢，我没有把这个想法扩展，题解就是将相邻的数量分组，比如 m个1，n个0，l个1. 结果就是min（m,n）+min(n,l).就是相邻求较小值。感觉想到了剪枝，但是没有把剪枝扩展开。

class Solution {
public:
    int countBinarySubstrings(string s) {
        int cnt = 0;
        vector<int> group;
        for(int i = 0; i< s.size(); i++){
            int j = i+1;
            while(s[j]==s[i])
                j++;
            // cout<<j-i<<endl;
            group.push_back(j-i);
            i = j - 1;
        }
        for(int i = 0; i < group.size()-1; i++)
            cnt += min(group[i],group[i+1]);
        return cnt;
    }
};