Find Smallest Letter Greater than Target
Problem
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’.
Note
lettershas a length in range[2, 10000].lettersconsists of lowercase letters, and contains at least 2 unique letters.targetis a lowercase letter.
Example
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
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"
My Answer
low,highindex를 이용해서 중간값mid를 찾고target보다 큰 문자 인지 확인- 만약
letters[mid]가target보다 크면,high에mid를 할당 - 만약
letters[mid]가target과 같거나 작으면,low에mid+1을 할당한다. letters[low]가 정답인데,letters[low]가target과 같거나 작은 경우가 발생할 수 있기 때문에 다음과 같이 보정해 준다.low를 하나 증가 시킨다.- 만약
low가letters.length와 같아지면low를0으로 할당해 준다.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
class Solution {
public char nextGreatestLetter(char[] letters, char target) {
int low=0;
int high=letters.length-1;
while(low < high) {
int mid = (low + high)/2;
if ( letters[mid] > target ) {
high = mid;
} else {
low = mid +1;
}
}
if ( letters[low] <= target ) {
low++;
if ( low >= letters.length ) {
low = 0;
}
}
return letters[low];
}
}