lc696. Count Binary Substrings

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原文链接：https://juejin.im/post/5cf61ee96fb9a07ec955fa72

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本文介绍了一种算法，用于计算字符串中具有相同数量连续0和1的非空子串的数量。通过记录连续出现的1和0的次数，算法能够准确地计算出满足条件的子串总数，包括重复子串的出现次数。

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：记录连续出现的1，0的次数，比如example，记录数组为[2,2,2,2],然后两两比较，返回较小的，[2,2,2],sum,得出结果

代码：python3

class Solution:
    def countBinarySubstrings(self, s: str) -> int:
        num=1
        arr=[]
        for i in range(1,len(s)):
            if s[i] == s[i-1]:
                num+=1
            else:
                arr.append(num)
                num=1
        arr.append(num)
        countArr=[min(arr[num],arr[num-1]) for num in range(1,len(arr))]
        return sum(countArr)
复制代码

转载于:https://juejin.im/post/5cf61ee96fb9a07ec955fa72