力扣题-12.6

[力扣刷题攻略] Re：从零开始的力扣刷题生活

力扣题1：696. 计数二进制子串

解题思想：先统计连续的0和连续的1的个数，然后进行相加即可（想不到一点）

class Solution(object):def countBinarySubstrings(self, s):""":type s: str:rtype: int"""count = []temp = s[0]num = 1for i in range(1,len(s)):if s[i]==temp:num +=1else:count.append(num)temp = s[i]num = 1count.append(num)result = 0for i in range(1,len(count)):result += min(count[i],count[i-1])return result

class Solution {
public:int countBinarySubstrings(string s) {std::vector<int> count;char temp = s[0];int num = 1;for (int i = 1; i < s.length(); ++i) {if (s[i] == temp) {num += 1;} else {count.push_back(num);temp = s[i];num = 1;}}count.push_back(num);int result = 0;for (int i = 1; i < count.size(); ++i) {result += std::min(count[i], count[i - 1]);}return result;        }
};