## 题目描述

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.

## 解法一

### 思路

``````  001110010    prev = 0    cur = 0
0 01110010    prev = 0    cur = 1
0 0 1110010    prev = 0    cur = 2
00 1 110010    prev = 2    cur = 1    01
001 1 10010    prev = 2    cur = 2    0011
0011 1 0010    prev = 2    cur = 3
00111 0 010    prev = 3    cur = 1    10
001110 0 10    prev = 3    cur = 2    1100
0011100 1 0    prev = 2    cur = 1    01
00111001 0     prev = 1    cur = 1    10``````

### Python

``````class Solution:
def countBinarySubstrings(self, s):
"""
:type s: str
:rtype: int
"""
result = 0
prev_length = 0
cur_length = 1
for i in range(1, len(s)):
if (s[i] == s[i - 1]):
cur_length += 1
else:
prev_length = cur_length
cur_length = 1
if prev_length >= cur_length:
result += 1
return result``````

### Java

``````class Solution {
public int countBinarySubstrings(String s) {
int result = 0;
int prevLength = 0;
int curLength = 1;
for (int i = 1; i < s.length(); i++) {
if (s.charAt(i) == s.charAt(i - 1)) {
curLength++;
} else {
prevLength = curLength;
curLength = 1;
}
if (prevLength >= curLength) {
result++;
}
}
return result;
}
}``````

### C++

``````class Solution {
public:
int countBinarySubstrings(string s) {
int result = 0;
int prevLength = 0;
int curLength = 1;
for (int i = 1; i < s.length(); i++) {
if (s[i] == s[i - 1]) {
curLength++;
} else {
prevLength = curLength;
curLength = 1;
}
if (prevLength >= curLength) {
result++;
}
}
return result;
}
};``````