# 910.Smallest Range II

**910.Smallest Range II**

> 给定一个整数数组 A，对于每个整数 A\[i]，我们可以选择 x = -K 或是 x = K，并将 x 加到 A\[i] 中。 在此过程之后，我们得到一些数组 B。 返回 B 的最大值和 B 的最小值之间可能存在的最小差值。

```
 示例 1：

 输入：A = [1], K = 0
 输出：0
 解释：B = [1]
 示例 2：

 输入：A = [0,10], K = 2
 输出：6
 解释：B = [2,8]
 示例 3：

 输入：A = [1,3,6], K = 3
 输出：3
 解释：B = [4,6,3]
  
提示：

  1 <= A.length <= 10000
  0 <= A[i] <= 10000
  0 <= K <= 10000
```

方法：可以先把数组排序，然后将数组看做两个部分，一部分是-K，一部分是+K。从头开始遍历，临界点之前的-K，临界点之后的+K，然后比较第一位+K和临界点右侧第一个-K的大小去最小值，以及最后一位-K后和临界点左边第一位+K后的大小取最大值，求出最大值与最小值的差。遍历一遍临界值，找出最小差值即可。

```
class Solution {
public:
  int smallestRangeII(vector<int>& A, int K) {
      sort(begin(A),end(A));
      int result=A.back()-A.front();
    
      for(int i=1;i<A.size();i++)
      {
         int l=min(A[0]+K,A[i]-K);
         int h=max(A.back()-K,A[i-1]+K);
          result = min(result,h-l);
      }
      return result;
  }
};
```


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