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

Approach #1: Math. [Java]

class Solution {
    public int smallestRangeI(int[] A, int K) {
        int min = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < A.length; ++i) {
            if (min > A[i]) min = A[i];
            if (max < A[i]) max = A[i];
        }
        if (max - min < 2 * K) return 0;
        return max - min - 2 * K;
    }
}

Analysis:

If you can find that the result only relate with (max - min) and 2 * K, it will become easy to solve.

908. Smallest Range I

原文：https://www.cnblogs.com/ruruozhenhao/p/10914741.html