leetcode 696. Count Binary Substrings

原题：

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.

代码如下：

int countBinarySubstrings(char* s) {
    int len = strlen(s);
    int result=0;
    int count=0;
    int temp=1;
    for(int n=0;ncount)
                result+=count;
            else
                result+=temp;
            count=temp;
            temp=1;
        }
    }
    if(temp>count)
        result+=count;
    else
        result+=temp;
    return result;
}

讲道理，真的是三天不练功，自己都看出来了。。。

真的是反应迟钝，，，

好吧 以后加强练习了。

这个基本的想法就是统计0，1连续的字符有多少，然后进行比较取小的就行。