例如: 输入: [1,2,3] 输出: 2 说明: 只有两个动作是必要的(记得每一步仅可使其中一个元素加1或减1): [1,2,3] => [2,2,3] => [2,2,2]
class Solution: def minMoves2(self, nums: List[int]) -> int: count = 0 if len(nums) == 1: return count nums.sort() for i in nums: count += abs(nums[len(nums)//2]-i) return count