原题链接在这里：https://leetcode.com/problems/count-binary-substrings/description/
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.lengthwill be between 1 and 50,000.swill only consist of "0" or "1" characters.
题解：
Count consecutive 0s or 1s.
e.g. 0001111, 就是3个0, 4个1.
When current pointing char is different. Add the min(preCount, curCount) to res.
Time Complexity: O(s.length()). Space: O(1).
AC Java:
1 class Solution { 2 public int countBinarySubstrings(String s) { 3 if(s == null || s.length() == 0){ 4 return 0; 5 } 6 7 int n = s.length(); 8 int preCount = 0; 9 int curCount = 0; 10 int res = 0; 11 12 int i = 0; 13 while(i<n){ 14 char cur = s.charAt(i); 15 while(i<n && s.charAt(i) == cur){ 16 curCount++; 17 i++; 18 } 19 20 res += Math.min(preCount, curCount); 21 preCount = curCount; 22 curCount = 0; 23 } 24 25 return res; 26 } 27 }