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
思路：找A数组的最大值和最小值，然后算差值，然后用+x去减少这个差值。明显答案就是看2K能不能覆盖这个差值呗
class Solution {
public:
int smallestRangeI(vector<int>& A, int K) {
int a = INT_MAX;
int b = INT_MIN;
for (int i = 0; i < A.size(); ++i) {
a = min(a, A[i]);
b = max(b, A[i]);
}
if (a == b) return 0;
if (b - a <= 2*K) return 0;
else {
return b - a - 2*K;
}
}
};