[Leetcode] 744. Find Smallest Letter Greater Than Target 解题报告

本文链接：https://blog.youkuaiyun.com/magicbean2/article/details/79525622

本文介绍了一个简单的算法问题，即在一个已排序的字符列表中找到第一个大于指定目标字母的元素。若列表中没有这样的元素，则返回列表中的第一个元素。文章通过示例详细解释了这一过程，并提供了一段C++代码实现。

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题目：

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.

思路：

由于letters是有序的，所有我们只需要简单地找到letters中第一个比target大的字符即可。如果不存在，则返回letters中的第一个元素即可。在C++中，upper_bound刚好就可以完成这个任务，其背后的原理就是二分查找。

代码：

class Solution {
public:
    char nextGreatestLetter(vector<char>& letters, char target) {
        auto it = upper_bound(letters.begin(), letters.end(), target);
        if (it == letters.end()) {
            return letters[0];
        }
        else {
            return *it;
        }
    }
};