題目：

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: s.length will be between 1 and 50,000. s will only consist of “0” or “1” characters.

解釋： 求字串中01個數一樣的substring的個數，注意substring是連續的哦。 參考：https://leetcode.com/problems/count-binary-substrings/discuss/108625/PythonC++Java-Easy-and-Concise-with-Explanation

python的奇技淫巧解法：

class Solution(object):
    def countBinarySubstrings(self, s):
        """
        :type s: str
        :rtype: int
        """
        lens=
 
map(len,s.replace('01','0 1').replace('10','1 0').split())
        return sum(min(a,b) for a,b in zip(lens,lens[1:])) 

對replace()的應用真是令人震驚，zip()函式在錯位計算的時候非常好用。 但是感覺這種解法還是不太常規的。 用c++實現以下：

class Solution {
public:
    int countBinarySubstrings(string s) {
        int cur=1,pre=0,res=0;
        for (int i=1;i<s.size();i++)
        {
            if(s[i]==s[i-1])
                cur++;
            else
            {
                res+=min(cur,pre);
                pre=cur;
                cur=1;
            }
        }
        res+=min(cur,pre);
        return res;
    }
};

c++的寫法真是妙不可言，更加常規，也比想象中的要清爽。 總結：