[leetcode-453-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]

 思路：

参考自：https://discuss.leetcode.com/topic/66737/it-is-a-math-question

let's define sum as the sum of all the numbers, before any moves; minNum as the min number int the list; n is the length of the list;

After, say m moves, we get all the numbers as x , and we will get the following equation

 sum + m * (n - 1) = x * n

and actually,

  x = minNum + m

and finally, we will get

  sum - minNum * n = m

So, it is clear and easy now.

int minMoves(vector<int>& nums)
     {
         int ret = 0;
         int smallest = *min_element(nums.begin(), nums.end());
         for (auto a:nums)
         {
             ret += (a - smallest);
         }
         return ret;
     }