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. 1 <= A.length <= 10000
2. 0 <= A[i] <= 10000
3. 0 <= K <= 10000

 

class Solution {
    public int smallestRangeI(int[] A, int K) {
        Arrays.sort(A);
        int dis = 0;
        dis=  A[A.length-1]- A[0];
        if(dis<2*K||dis == 2*K) return 0;
        else return  (dis - 2*K);
    }
}

这道题很简单，但是很容易写成上述代码的形式。时间复杂度最坏O(n^2)，最好O(nlogn)。

找最大值和最小值，O(n)时间足够了。题目虽然容易，但是菜鸟和大神之间的差距就是从这么简单的题目开始的。

 

class Solution {
    public int smallestRangeI(int[] A, int K) {
         int min = A[0];
         int max = A[0];
         for(int a : A){
             if(min > a) min = a;
             if(max < a) max = a;
         }
        if(max - min <2*K || max - min == 2*K) return 0;
        else return (max-min-2*K);
      
    }
}

时间复杂度为O(n），Runtime: 5 ms, faster than 96.45% of Java online submissions for Smallest Range I.