给你一个排序后的字符列表 letters
,列表中只包含小写英文字母。另给出一个目标字母 target
,请你寻找在这一有序列表里比目标字母大的最小字母。
在比较时,字母是依序循环出现的。举个例子:
- 如果目标字母
target = 'z'
并且字符列表为letters = ['a', 'b']
,则答案返回'a'
示例:
输入: letters = ["c", "f", "j"] target = "a" 输出: "c" 输入: letters = ["c", "f", "j"] target = "c" 输出: "f" 输入: letters = ["c", "f", "j"] target = "d" 输出: "f" 输入: letters = ["c", "f", "j"] target = "g" 输出: "j" 输入: letters = ["c", "f", "j"] target = "j" 输出: "c" 输入: letters = ["c", "f", "j"] target = "k" 输出: "c"
提示:
letters
长度范围在[2, 10000]
区间内。letters
仅由小写字母组成,最少包含两个不同的字母。- 目标字母
target
是一个小写字母。
class Solution:
def nextGreatestLetter(self, letters: List[str], target: str) -> str:
left, right = 0, len(letters)
while left < right:
mid = (left + right) >> 1
if ord(letters[mid]) > ord(target):
right = mid
else:
left = mid + 1
return letters[left % len(letters)]
class Solution {
public char nextGreatestLetter(char[] letters, char target) {
int left = 0, right = letters.length;
while (left < right) {
int mid = (left + right) >> 1;
if (letters[mid] > target) {
right = mid;
} else {
left = mid + 1;
}
}
return letters[left % letters.length];
}
}
function nextGreatestLetter(letters: string[], target: string): string {
let left = 0, right = letters.length;
let x = target.charCodeAt(0);
while (left < right) {
let mid = (left + right) >> 1;
if (x < letters[mid].charCodeAt(0)) {
right = mid;
} else {
left = mid + 1;
}
}
return letters[left % letters.length];
};
class Solution {
public:
char nextGreatestLetter(vector<char>& letters, char target) {
int left = 0, right = letters.size();
while (left < right) {
int mid = left + right >> 1;
if (letters[mid] > target) {
right = mid;
} else {
left = mid + 1;
}
}
return letters[left % letters.size()];
}
};
func nextGreatestLetter(letters []byte, target byte) byte {
left, right := 0, len(letters)
for left < right {
mid := (left + right) >> 1
if letters[mid] > target {
right = mid
} else {
left = mid + 1
}
}
return letters[left%len(letters)]
}