题目

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

链接

https://leetcode.com/problems/smallest-range-i/

分析

A的最大值和最小值之间的diff，如果在[0, 2K]之间，那么一定可以调成一个所有元素都相等的数组；
否则可想办法消去这个2K的diff

代码

class Solution(object):
    """
    A的最大值和最小值之间的diff，如果在[0, 2K]之间，那么一定可以调成一个所有元素都相等的数组；
    否则可想办法消去这个2K的diff
    """
    def smallestRangeI(self, A, K):
        """
        :type A: List[int]
        :type K: int
        :rtype: int
        """
        min = 10000
        max = 0
        for ele in A:
            if ele < min:
                min = ele
            if ele > max:
                max = ele
        diff = max - min
        if diff >= 0 and diff < 2*K:
            return 0
        else:
            return diff - 2*K