744. Find Smallest Letter Greater Than Target*

744. Find Smallest Letter Greater Than Target*

https://leetcode.com/problems/find-smallest-letter-greater-than-target/

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. 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.

C++ 实现 1

在 letters 中查找 target 的 upper_bound, 找到了, 则返回; 没找到, 则返回 letters[0].

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

C++ 实现 2

也可以这样写;

class Solution {
public:
    char nextGreatestLetter(vector<char>& letters, char target) {
        int l = 0, r = letters.size() - 1;
        while (l <= r) {
            int mid = l + (r - l) / 2;
            if (letters[mid] <= target)
                l = mid + 1;
            else
                r = mid - 1;
        }
        if (l >= 0 && l < letters.size())
            return letters[l];
        return letters[0];
    }
};