leetcode 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: 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.

思路就是，先找出最小的满足要求的单元比如01或10这两种结构。
然后在找到的这些单元中进行扩展。直到结束。
时间复杂度，不太会算，但是应该不会太慢

class Solution {
public:
    int countBinarySubstrings(string s) {
        int n = s.size();
        vector<pair<int,int> > v;
        int num = 0;
        int ans = 0;
        for (int i = 1; i < n; ++i) {
            if (s[i] != s[i-1]) v.push_back({i-1, i}), ++ans;
        }
        while (true) {
            int mark = 0;
            for (int i = 0; i < v.size(); ++i) {
                while(v[i].first-1 >=0 && v[i].second+1 < n && s[v[i].first-1] == s[v[i].first] && s[v[i].second+1]==s[v[i].second]){
                    ans++;
                    v[i].first--;
                    v[i].second++;
                    mark=1;
                }
            }
            if (!mark) break;
        }
        return ans;
    }
};