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 <= 100000 <= A[i] <= 100000 <= K <= 10000int smallestRangeI(vector<int>& A, int K) { int minV = A[0], maxV = A[0]; for(int i = 0; i < A.size(); i++) { minV = min(minV, A[i]); maxV = max(maxV, A[i]); } return max(((maxV - minV) - 2 * K),0); }
[leetcode-908-Smallest Range I]
原文：https://www.cnblogs.com/hellowooorld/p/9749846.html