Give a strings, 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.

这道题的大概思路是现在找到01或者10这种子字符串，然后向两边扩散，找符合条件的子字符串

class Solution {
    public int countBinarySubstrings(String s) {
        if(s == null || s.length() <= 1)
            return 0;

        int res = 0;
        for(int i = 0; i < s.length() - 1; i++) {
            if(s.charAt(i) != s.charAt(i + 1)) {
                res++;
                int left = i - 1;
                int right = i + 2;

                while(left >= 0 && right < s.length() && s.charAt(left) == s.charAt(left + 1) && s.charAt(right) == s.charAt(right - 1)) {
                    res++;
                    left--;
                    right++;
                }
            }
        }

        return res;
    }
}