地址：https://leetcode.com/problems/count-binary-substrings/

题目：

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.

理解：

如何计算多少个子串中包含相同位数且连续的0和1？
比如111000这种形式，就包括了111000，1100，10三种。也就是说，当遇到一个01翻转的时候，就可以判断前面的位置中子串的个数了。取连0和连1中的较小值。

实现：

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