Approach

# 搜尋陣列找比target大的字元

Time Complexity

O(n)

Space Complexity

O(1)

Code

class Solution {
public:
    char nextGreatestLetter(vector<char>& letters, char target) {
        int where = -1;

        for (int i = 0 ; i < letters.size() ; i++) {
            if (letters[i] - target > 0) {
                where = i;
                break;
            }
        }

        if (where == -1) {
            return letters[0];
        }
        else {
            return letters[where];
        }
    }
};