[LeetCode] Find Smallest Letter Greater Than Target

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.

找出比目标值大的最小的那个字母

给定一个字符数组，找出比目标值大的最小的那个字符，如果目标值大于数组中最大的字母，则返回数组中第一个字母。

使用upper_bound()这个函数找出比目标字母大的那个数，这时需要判断

如果目标值大于等于数组中最大的字母，则返回数组中的首字母。

否则返回找到的那个字母。

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.front();
        else {
            return *it;
        }
    }
};
// 12 ms

也可以使用二分搜索。

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