744 Find Smallest Letter Greater Than Target

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本文介绍了一个算法问题：在一个已排序的小写字母列表中找到第一个大于指定目标字母的元素。若目标字母大于列表中所有字母，则返回列表的第一个元素。文章通过示例展示了该算法的工作原理。

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.

class Solution {
  public char nextGreatestLetter(char[] letters, char target) {
    if (letters.length < 1) { return target; }
    if (letters[letters.length - 1] < target) { return letters[0]; }
    for (int i = 0; i < letters.length; i++) {
      if (target < letters[i]) {
        return letters[i];
      }
    }
    return letters[0];
  }
}