[leetcode-908-Smallest Range I]

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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 = 0Output: 0Explanation: B = [1]

Example 2:

Input: A = [0,10], K = 2Output: 6Explanation: B = [2,8]

Example 3:

Input: A = [1,3,6], K = 3Output: 0Explanation: 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:很樸素的想法,掃描一趟即可知道最大值與最小值,然後再比較一下即可。
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);}

 

[leetcode-908-Smallest Range I]

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