Minimum Moves to Equal Array Elements

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本文探讨了如何通过最少的步骤使数组中的所有元素相等。每次操作可以选择递增数组中除一个元素外的所有元素。文章提供了两种算法实现，一种是通过排序后逐个比较并累加差值，另一种则是利用总和减去最小值乘以数组长度来快速得出结果。

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Given a non-empty integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1.

Example:

Input:
[1,2,3]

Output:
3

Explanation:
Only three moves are needed (remember each move increments two elements):

[1,2,3]  =>  [2,3,3]  =>  [3,4,3]  =>  [4,4,4]

class Solution(object):
    def minMoves(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        nums.sort()
        c = 0
        for i in range(len(nums)-1, -1, -1):
            if nums[i] == nums[0]:
                break
            c += nums[i] - nums[0]
            
        return c

class Solution(object):
    def minMoves(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return sum(nums) - len(nums)*min(nums)