744 Find Smallest Letter Greater Than Target
1. Question
Given a list of sorted charactersletters
containing only lowercase letters, and given a target lettertarget
, find the smallest element in the list that is larger than the given target.
Letters also wrap around. For example, if the target istarget = 'z'
andletters = ['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.
2. Implementation
(1) Binary Search
思路: 非常典型的二分查找 找左边界,思路和search insert position是一样的,其中注意题目要求letters是可以wrap around的 (也就是如果target在数组的插入点是在数组末端的时候,则数组的第一个字母为所求解)
class Solution {
public char nextGreatestLetter(char[] letters, char target) {
int start = 0, end = letters.length - 1, mid = 0;
while (start + 1 < end) {
mid = start + (end - start) / 2;
if (letters[mid] <= target) {
start = mid + 1;
}
else {
end = mid;
}
}
if (letters[start] > target) {
return letters[start];
}
else if (letters[end] > target) {
return letters[end];
}
else {
return letters[0];
}
}
}
3. Time & Space Complexity
Binary Search: 时间复杂度O(logn), 空间复杂度O(1)
Last updated
Was this helpful?