LeetCode | 1374. Generate a Stri

LeetCode 1374. Generate a String With Characters That Have Odd Counts生成每种字符都是奇数个的字符串【Easy】【Python】【字符串】

Problem

LeetCode

Given an integer n, return a string with n characters such that each character in such string occurs an odd number of times.

The returned string must contain only lowercase English letters. If there are multiples valid strings, return any of them.

Example 1:

Input: n = 4
Output: "pppz"
Explanation: "pppz" is a valid string since the character 'p' occurs three times and the character 'z' occurs once. Note that there are many other valid strings such as "ohhh" and "love".

Example 2:

Input: n = 2
Output: "xy"
Explanation: "xy" is a valid string since the characters 'x' and 'y' occur once. Note that there are many other valid strings such as "ag" and "ur".

Example 3:

Input: n = 7
Output: "holasss"

Constraints:

• 1 <= n <= 500

问题

力扣

给你一个整数 n，请你返回一个含 n 个字符的字符串，其中每种字符在该字符串中都恰好出现 奇数次 。

返回的字符串必须只含小写英文字母。如果存在多个满足题目要求的字符串，则返回其中任意一个即可。

示例 1：

输入：n = 4
输出："pppz"
解释："pppz" 是一个满足题目要求的字符串，因为 'p' 出现 3 次，且 'z' 出现 1 次。当然，还有很多其他字符串也满足题目要求，比如："ohhh" 和 "love"。

示例 2：

输入：n = 2
输出："xy"
解释："xy" 是一个满足题目要求的字符串，因为 'x' 和 'y' 各出现 1 次。当然，还有很多其他字符串也满足题目要求，比如："ag" 和 "ur"。

示例 3：

输入：n = 7
输出："holasss"

提示：

• 1 <= n <= 500

思路

字符串

n 为偶数，n-1 个 'a' + 1 个 'b'
n 为奇数，n 个 'a'

时间复杂度: O(n)
空间复杂度: O(n)

Python3代码

class Solution:
    def generateTheString(self, n: int) -> str:
        res = []
        if n % 2 == 0:
            for i in range(n-1):
                res.append('a')
            res.append('b')
        else:
            for i in range(n):
                res.append('a')
        return ''.join(res)

代码地址

GitHub链接