696. Count Binary Substrings

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: 6Explanation: 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: 4Explanation: There are 4 substrings: "10", "01", "10", "01" that have equal number of consecutive 1's and 0's.

此题主要需要明白问题核心可以简化问题 ，观察1100、11100、1100000，可以看到零一配对的一组数，根据题目规则能得到的个数受零或一的个数中最小的影响，即以上三个数组都是2.因此问题转化为，取两个相邻连续零或一个数中最小的值加到最后的总和中。

那么循环设置为遍历数组，与第一个数字相同值时计数，遇到不同值时开辟新的计数，再下次遇到与第二次的不同值时选出最小的计数加到和中。将第二次计数赋给第一个计数器，重复接下来的步骤。

class Solution {public:    int countBinarySubstrings(string s) {                int size = s.size();        if(size==0) return 0;        char now=s[0];        int l = 1;        int ll = 0;        int sum=0;        for(int i=1;i<size;i++){            if(s[i]==now){                if(ll!=0){                    sum += (l>ll? ll:l);                    now = s[i-1];                    l=ll;                    ll=1;                    continue;                }                l++;            }            else{                ll++;            }        }        sum += (l>ll? ll:l);        return sum;    }};