- Description:
- 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
- Accepted
- 29,142
- Submissions
- 44,849
- Solution:
- class Solution {
- public int smallestRangeI(int[] A, int K) {
- if(A==null||A.length==1){
- return 0;
- }
- Arrays.sort(A);
- int second = A[0];
- int first = A[A.length -1];
- if((first - second)<=2*K){
- return 0 ;
- }
- else {
- return first - second - K - K ;
- }
- }
- }
来源: http://www.bubuko.com/infodetail-3209626.html