696. Count Binary Substrings -- Python

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本文介绍了一种算法，用于计算给定字符串中具有相同数量连续0和1的非空子串的数量，并通过示例解释了如何实现该算法。

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：
可以先将连续的 ”0“ 和 ”1“ 分离，然后再依次找个两个中的最小值就可

‘00001111’ => [4, 4] => min(4, 4) => 4
‘00110’ => [2, 2, 1] => min(2, 2) + min(2, 1) => 3
‘10101’ => [1, 1, 1, 1, 1] => 4

我的代码：（效率太低，超时，未被接受）：

class Solution:
    def countBinarySubstrings(self, nums):
        """
        :type s: str
        :rtype: int
        """
        number=0
        nums_of_0 = nums.count("0")
        nums_of_1 = nums.count("1")
        for i in range(1,min(nums_of_0,nums_of_1)+1):
            x=["0"*i+"1"*i,"1"*i+'0'*i]
            count1=nums.count(x[0])
            count2=nums.count(x[1])
            number+=count1+count2
        return number

修改版：

class Solution:
    def countBinarySubstrings(self, s):
        """
        :type s: str
        :rtype: int
        """
        l=[]
        for i in s.replace("01","0 1").replace("10","1 0").split():
            l.append(len(i))
        return sum(min(a,b) for a,b in zip(l,l[1:]))