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[];
int n = letters.size(), left = , right = n;
while (left < right) {
int mid = left + (right - left) / ;
if (letters[mid] <= target) left = mid + ;
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;
}
};