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]
思路：
很明显答案就是容器里最大值与最小值的差值，如果这个值小于2K，那么结果就为0，当然需要注意只有一个元素的情况。这才是名副其实的EZ啊。
Solution:
1 class Solution { 2 public: 3 int smallestRangeI(vector<int>& A, int K) 4 { 5 sort(A.begin(),A.end()); 6 int size = A.size(); 7 int result = 0; 8 if(size>1) 9 { 10 result = A[size-1] - A[0]; 11 } 12 result -=2*K; 13 if(result<0) 14 { 15 result = 0; 16 } 17 return result; 18 } 19 };
直接调用stl自带的sort排序，简直不用太爽。哈哈哈