LeetCode | 0744. Find Smallest Letter Greater Than Target寻找比目标字母大的最小字母【Python】

LeetCode 0744. Find Smallest Letter Greater Than Target寻找比目标字母大的最小字母【Easy】【Python】【二分】

Problem

LeetCode

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

问题

力扣

给定一个只包含小写字母的有序数组 letters 和一个目标字母 target，寻找有序数组里面比目标字母大的最小字母。

数组里字母的顺序是循环的。举个例子，如果目标字母 target = 'z' 并且有序数组为 letters = ['a', 'b']，则答案返回 'a'。

示例:

输入:
letters = ["c", "f", "j"]
target = "a"
输出: "c"

输入:
letters = ["c", "f", "j"]
target = "c"
输出: "f"

输入:
letters = ["c", "f", "j"]
target = "d"
输出: "f"

输入:
letters = ["c", "f", "j"]
target = "g"
输出: "j"

输入:
letters = ["c", "f", "j"]
target = "j"
输出: "c"

输入:
letters = ["c", "f", "j"]
target = "k"
输出: "c"

注:

1. letters 长度范围在 [2, 10000] 区间内。
2. letters 仅由小写字母组成，最少包含两个不同的字母。
3. 目标字母 target 是一个小写字母。

思路

二分查找

注意数组是循环的，所以如果 target >= 最后一个字母，直接返回 letters[0] 即可。

时间复杂度: O(logn)
空间复杂度: O(1)

Python代码

class Solution(object):
    def nextGreatestLetter(self, letters, target):
        """
        :type letters: List[str]
        :type target: str
        :rtype: str
        """
        low, high = 0, len(letters) - 1
        while low <= high:
            mid = int((low + high) / 2)  # element in list must be int
            if letters[mid] <= target:
                low = mid + 1
            else:
                if mid < 1 or (mid >= 1 and letters[mid-1] <= target):
                    return letters[mid]
                high = mid - 1
        return letters[0]  # 'z' < 'a'

代码地址

GitHub链接