【leetcode】1525. Number of Good Ways to Split a String

题目如下:

You are given a string s, a split is called good if you can split s into 2 non-empty strings p and q where its concatenation is equal to s and the number of distinct letters in p and q are the same.

Return the number of good splits you can make in s.

Example 1:

Input: s = "aacaba"
Output: 2
Explanation: There are 5 ways to split "aacaba" and 2 of them are good. 
("a", "acaba") Left string and right string contains 1 and 3 different letters respectively.
("aa", "caba") Left string and right string contains 1 and 3 different letters respectively.
("aac", "aba") Left string and right string contains 2 and 2 different letters respectively (good split).
("aaca", "ba") Left string and right string contains 2 and 2 different letters respectively (good split).
("aacab", "a") Left string and right string contains 3 and 1 different letters respectively.

Example 2:

Input: s = "abcd"
Output: 1
Explanation: Split the string as follows ("ab", "cd").

Example 3:

Input: s = "aaaaa"
Output: 4
Explanation: All possible splits are good.

Example 4:

Input: s = "acbadbaada"
Output: 2 

Constraints:

  • s contains only lowercase English letters.
  • 1 <= s.length <= 10^5

解题思路:本题不难,首先统计出s中所有字符出现的次数,然后再从左往右遍历s,依次减少之前统计的结果即可。

代码如下:

class Solution(object):
    def numSplits(self, s):
        """
        :type s: str
        :rtype: int
        """
        res = 0
        dic_right = {}
        dic_left = {}
        for i in s:
            dic_right[i] = dic_right.setdefault(i,0) + 1
        for i in s:
            dic_left[i] = dic_left.setdefault(i,0)+1
            dic_right[i] -= 1
            if dic_right[i] == 0:
                del dic_right[i]
            if len(dic_left) == len(dic_right):
                res += 1

        return res

 

posted @ 2020-09-14 15:20  seyjs  阅读(381)  评论(0编辑  收藏  举报