Find Smallest Letter Greater Than Target

犯的错：退出循环后 check start 和end 应该check > 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.

注意 找不到时，返回第一个字母

时间复杂度 o(logn)

class Solution {
    public char nextGreatestLetter(char[] letters, char target) {
        int start = 0, end = letters.length - 1;
        
        while(start + 1 < end){
            int mid = start + (end - start)/2;
            
            if(letters[mid] <= target)
                start = mid;
            else
                end = mid;
        }
        //不能等于，不需要找到target，而是要找到比target大的最小的
        if(letters[start] > target){
            return letters[start];
        }
        
        if(letters[end] > target){
            return letters[ end];
        }
        
        return letters[0];
        
    }
}

class Solution {
    public char nextGreatestLetter(char[] letters, char target) {
        if(letters == null || letters.length == 0) return 0;
        
        int low = 0, high = letters.length -1;
        
        while(low <= high){
            int mid = low + ((high-low)>>1);
            
            if(letters[mid] > target){
                if(mid == 0 || letters[mid-1] <= target) return letters[mid];
                else
                    high = mid -1;
                
            }else{
                low = mid + 1;
            }
        }
        
        return letters[0];
    }
}