[leetcode-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. 1 <= A.length <= 10000
2. 0 <= A[i] <= 10000
3. 0 <= K <= 10000

 思路：

很朴素的想法，扫描一趟即可知道最大值与最小值，然后再比较一下即可。

int 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);
}