forked from fishercoder1534/Leetcode
-
Notifications
You must be signed in to change notification settings - Fork 0
/
_908.java
67 lines (62 loc) · 1.4 KB
/
_908.java
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
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
package com.fishercoder.solutions;
import java.util.Arrays;
/**
* 908. Smallest Range I
*
* Given an array A of integers, for each integer A[i] we may choose any x with -K <= x <= K, and add x to A[i].
*
* After this process, we have some array B.
*
* Return the smallest possible difference between the maximum value of B and the minimum value of B.
*
*
*
* Example 1:
*
* Input: A = [1], K = 0
* Output: 0
* Explanation: B = [1]
* Example 2:
*
* Input: A = [0,10], K = 2
* Output: 6
* Explanation: B = [2,8]
* Example 3:
*
* Input: A = [1,3,6], K = 3
* Output: 0
* Explanation: B = [3,3,3] or B = [4,4,4]
*
*
* Note:
*
* 1 <= A.length <= 10000
* 0 <= A[i] <= 10000
* 0 <= K <= 10000
*/
public class _908 {
public static class Solution1 {
public int smallestRangeI(int[] A, int K) {
Arrays.sort(A);
int smallestPlus = A[0] + K;
int biggestMinus = A[A.length - 1] - K;
int diff = biggestMinus - smallestPlus;
if (diff > 0) {
return diff;
} else {
return 0;
}
}
}
public static class Solution2 {
public int smallestRangeI(int[] A, int K) {
int min = A[0];
int max = A[0];
for (int k : A) {
min = Math.min(min, k);
max = Math.max(max, k);
}
return Math.max(max - min - 2 * K, 0);
}
}
}