<h3><a title="题目描述" href="https://leetcode-cn.com/problems/count-binary-substrings/">题目描述</a></h3>  <h4>方法一：按连续相同字符计数分组</h4> <pre class="EnlighterJSRAW" data-enlighter-language="python">class Solution: def countBinarySubstrings(self,s): counts = [] i = 0 while i < len(s): j = i while j < len(s) and s[i] == s[j]: j += 1 counts.append(j-i) i = j ans = 0 for i in range(len(counts)-1): ans += min(counts[i],counts[i+1]) return ans </pre> <p> </p> <h4>方法二</h4> <pre class="EnlighterJSRAW" data-enlighter-language="python">class Solution: def countBinarySubstrings(self,s): cur,last,res = 1,0,0 for i in range(1,len(s)): if(s[i] == s[i-1]): cur += 1 else: last = cur cur = 1 if(last >= cur): res += 1 return res </pre> <p> </p> ## 思路： - 方法一 ： 具体可见 [题解](https://leetcode-cn.com/problems/count-binary-substrings/solution/ji-shu-er-jin-zhi-zi-chuan-by-leetcode-solution/ "题解") - 方法二 ：用last来记录之前一种数字的个数， cur来记录当前数字的个数； 当last >= cur的时候， res ++;