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:
1、 lettershasalengthinrange[2,10000].；
2、 lettersconsistsoflowercaseletters,andcontainsatleast2uniqueletters.；
3、 targetisalowercaseletter.；
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]
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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)]
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