Minimum Moves to Equal Array Elements

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]

 

实际上，每次更新需要最小的n-1个值去更新。假设总共更新k次。

那么，最终的情况，所有数的和等于(min+k) * n，min是初始数据的最小值，n是数组的长度；

更新k次，每次更新n-1个值，那么就有 sum+(n-1)*k = （min+k)*n；

化简一下，k=sum-min*n

class Solution {
public:
    int minMoves(vector<int>& nums) {
        int min = INT_MAX, sum = 0;
        for (auto n : nums) {
            if (n < min) min = n;
            sum += n;
        }
        return sum - min * nums.size();
    }
};