[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 <= A.length <= 10000
0 <= A[i] <= 10000
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[], maxV = A[];

    for(int i = ; i < A.size(); i++)

    {

        minV = min(minV, A[i]);

        maxV = max(maxV, A[i]);

    }

    return max(((maxV - minV) -  * K),);

}

巴特西

[leetcode-908-Smallest Range I]

最新文章

热门文章