QUESTION:

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.

EXPLANATION:

首先需要审题的是：1.已经是一个顺序数组了。2.并且是首尾相连的数组。 那么就可以这么考虑：既然首尾相连，那么我们就直接使用两个数组就好了。 所以简化成： 1.将所有字符放在数组中，同时在+26的位置也添加上 2.寻找target后的第一个字符 3.返回就可以

SOLUTION：

class Solution {
    public char nextGreatestLetter(char[] letters, char target) {
        int[] index = new int[52];
        for(char c:letters){
            index[c-'a'] +=1;
            index[c-'a'+26] +=1;
        }
        int cursor = target-'a'+1;
        while (index[cursor]==0)
            cursor++;
        return (char)('a'+cursor%26);
    }
}