744. 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.

二刷：

如果没有比target大的元素，就返回数组第一个元素，time: O(log(n)), space: O(1)

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

or

class Solution {
    public char nextGreatestLetter(char[] letters, char target) {
        int left = 0, right = letters.length - 1;
        while(left <= right) {
            int mid = left + (right - left) / 2;
            if(letters[mid] <= target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        return letters[left % letters.length];
    }
}

一刷：

如果没有比target大的元素，就返回数组第一个元素

time: O(log(n)), space: O(1)

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