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.
这道题给了我们一堆有序的字母,然后又给了我们一个target字母,让我们求字母数组中第一个大于target的字母,数组是循环的,如果没有,那就返回第一个字母。像这种在有序数组中找数字,二分法简直不要太适合啊。题目中说了数组至少有两个元素,那么我们首先用数组的尾元素来跟target比较,如果target大于等于尾元素的话,直接返回数组的首元素即可。否则就利用二分法来做,这里是查找第一个大于目标值的数组,博主之前做过二分法的总结,参见这个帖子LeetCode Binary Search Summary 二分搜索法小结,参见代码如下:
解法一:
class Solution { public: char nextGreatestLetter(vector<char>& letters, char target) { if (target >= letters.back()) return letters[0]; int n = letters.size(), left = 0, right = n; while (left < right) { int mid = left + (right - left) / 2; if (letters[mid] <= target) left = mid + 1; else right = mid; } return letters[right]; } };
我们也可以用STL自带的upper_bound函数来做,这个就是找第一个大于目标值的数字,如果返回end(),说明没找到,返回首元素即可,参见代码如下:
解法二:
class Solution { public: char nextGreatestLetter(vector<char>& letters, char target) { auto it = upper_bound(letters.begin(), letters.end(), target); return it == letters.end() ? *letters.begin() : *it; } };