[LeetCode] Find Smallest Letter Greater Than Target

1.문제

You are given an array of characters letters that is sorted in non-decreasing order, and a character target. There are at least two different characters in letters.

Return the smallest character in letters that is lexicographically greater than target. If such a character does not exist, return the first character in letters.

영어 알파벳이 오름차순으로 정렬되어있는 배열이 주어지고 특정 영어 알파벳인 target이 주어질 때 letters에서 사전 순서에서 target보다 큰 철자들 중 가장 사전 순서로 앞에 있는 철자를 리턴하는 문제이다. 만약 target보다 사전 순서가 뒤 인 알파벳이 없다면 letters[0]을 리턴하면 된다.

Example 1

Input: letters = ["c","f","j"], target = "a"
Output: "c"
Explanation: The smallest character that is lexicographically greater than 'a' in letters is 'c'.

Example 2

Input: letters = ["c","f","j"], target = "c"
Output: "f"
Explanation: The smallest character that is lexicographically greater than 'c' in letters is 'f'.

Example 3

Input: letters = ["x","x","y","y"], target = "z"
Output: "x"
Explanation: There are no characters in letters that is lexicographically greater than 'z' so we return letters[0].

Constraints:

• 2 <= letters.length <= 10^4
• letters[i] is a lowercase English letter.
• letters is sorted in non-decreasing order.
• letters contains at least two different characters.
• target is a lowercase English letter.

2.풀이

1. letters를 반복 하면서 target보다 사전 순서가 뒤인 알파벳인지 확인한다.

/**
 * @param {character[]} letters
 * @param {character} target
 * @return {character}
 */
const nextGreatestLetter = function (letters, target) {
  let result = letters[0];

  for (let i = 0; i < letters.length; i++) {
    // target 보다 사전 순서가 크다면 result 에 저장 후 break
    if (letters[i] > target) {
      result = letters[i];

      break;
    }
  }

  return result;
};

3.결과