描述
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
解析
根据题意，A 中的任何一个数字和 -K <= x <= K 中的任何一个数字相加，得到数组 B ，然后求 min(max(B)-min(B))，只要把所有的数字向中间靠拢。即 A 中的最小值 +K , 最大值 -K ，判断这样操作之后，能否有重叠，如果能重叠所求的结果就是 0；如果不能重叠，所求的结果就是两者的差值。时间复杂度为 O(N)，空间复杂度为 O(1)。
解答
class Solution(object):
def smallestRangeI(self, A, K):
"""
:type A: List[int]
:type K: int
:rtype: int
"""
return max(max(A) - min(A) - 2 * K, 0)
运行结果
Runtime: 96 ms, faster than 84.36% of Python online submissions for Smallest Range I.
Memory Usage: 12.7 MB, less than 32.29% of Python online submissions for Smallest Range I.
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