Formatted question description: https://leetcode.ca/all/744.html
744. Find Smallest Letter Greater Than Target (Easy)
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.
Solution 1.
// OJ: https://leetcode.com/problems/find-smallest-letter-greater-than-target/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
char nextGreatestLetter(vector<char>& letters, char target) {
auto i = upper_bound(letters.begin(), letters.end(), target);
return i == letters.end() ? letters[0] : *i;
}
};
Java
class Solution {
public char nextGreatestLetter(char[] letters, char target) {
int length = letters.length;
char lastLetter = letters[length - 1];
if (lastLetter <= target)
return letters[0];
int index = binarySearch(letters, target);
return letters[index];
}
public int binarySearch(char[] letters, char target) {
target++;
int low = 0, high = letters.length - 1;
while (low <= high) {
int mid = (high - low) / 2 + low;
char c = letters[mid];
if (c == target)
return mid;
else if (c > target)
high = mid - 1;
else
low = mid + 1;
}
return low;
}
}