# 908.Smallest Range I

**908.Smallest Range I**

> 给定一个整数数组 A，对于每个整数 A\[i]，我们可以选择任意 x 满足 -K <= 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：0
 解释：B = [3,3,3] 或 B = [4,4,4]

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

方法：这题比较简单，直接比较最大值-K和最小值+K后的大小，如果小于则返回0，大于返回差值。

```
class Solution {
public:
  int smallestRangeI(vector<int>& A, int K) {
      int max=A[0],min=A[0];
      for(int i=1;i<A.size();i++)
      {
          if(max<A[i]) max=A[i];
          if(min >A[i]) min=A[i];
      }

       max -=K;
      min +=K;
      if(max <=min) return 0;
      return max-min;
      
  }
};
```


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