【LeetCode】744. Find Smallest Letter Greater Than Target 解题报告（Python）

作者：负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

题目描述

Given a list of sorted characters letters containing only lowercase letters, and given a target letter target, find the smallest element in the list that is larger than the given target.

Letters also wrap around. For example, if the target is target = 'z' and letters = ['a', 'b'], the answer is 'a'.

Examples:

Input:

letters = ["c", "f", "j"]

target = "a"

Output: "c"

Input:

letters = ["c", "f", "j"]

target = "c"

Output: "f"

Input:

letters = ["c", "f", "j"]

target = "d"

Output: "f"

Input:

letters = ["c", "f", "j"]

target = "g"

Output: "j"

Input:

letters = ["c", "f", "j"]

target = "j"

Output: "c"

Input:

letters = ["c", "f", "j"]

target = "k"

Output: "c"

Note:

letters has a length in range [2, 10000].
letters consists of lowercase letters, and contains at least 2 unique letters.
target is a lowercase letter.

题目大意

给出了一个排序的字符数组，找出第一个比target大的字符。注意是数组是循环的。

解题方法

线性扫描

找到第一个比指定字符大的。如果没找到的话，列表是可以循环的。也就是说如果没找到就返回列表第一个字符。

class Solution(object):

    def nextGreatestLetter(self, letters, target):

        """

        :type letters: List[str]

        :type target: str

        :rtype: str

        """

        for letter in letters:

        ## 提交了之后发现不用使用ord，字符可以用'>''<'比较大小

            if ord(letter) > ord(target):

                return letter

        return letters[0]

二分查找

因为是有序的，所以直接二分即可。

class Solution(object):

    def nextGreatestLetter(self, letters, target):

        """

        :type letters: List[str]

        :type target: str

        :rtype: str

        """

        index = bisect.bisect_right(letters, target)

        return letters[index % len(letters)]

日期

2018 年 1 月 23 日
2018 年 11 月 19 日 —— 周一又开始了

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