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.
找出比目标值大的最小的那个字母
给定一个字符数组,找出比目标值大的最小的那个字符,如果目标值大于数组中最大的字母,则返回数组中第一个字母。
使用upper_bound()这个函数找出比目标字母大的那个数,这时需要判断
如果目标值大于等于数组中最大的字母,则返回数组中的首字母。
否则返回找到的那个字母。
class Solution { public: char nextGreatestLetter(vector<char>& letters, char target) { auto it = upper_bound(letters.begin(), letters.end(), target); if (it == letters.end()) return letters.front(); else { return *it; } } }; // 12 ms
也可以使用二分搜索。
class Solution { public: char nextGreatestLetter(vector<char>& letters, char target) { if (letters.back() <= target) return letters.front(); int left = 0, right = letters.size() - 1; while (left < right) { int mid = left + (right - left) / 2; if (letters[mid] > target) { right = mid; } else { left = mid + 1; } } return letters[left]; } };