LeetCode-Minimum_Moves_to_Equal_Array_Elements

题目：

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]

翻译:

给定一个大小为 n 非空的数组，找到最小的移动次数使得所有数组中元素大小相等，每移动一次，使 n - 1 个元素增加1。

思路：

把这道题看成一道数学题去解决，解析过程参见如下，摘自leetCode中的解答。

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

This part may be a little confusing, but @shijungg explained very well. let me explain a little again. it comes from two observations:

1. the minum number will always be minum until it reachs the final number, because every move, other numbers (besides the max) will be increamented too;
2. from above, we can get, the minum number will be incremented in every move. So, if the final number is x, it would be minNum + moves;

and finally, we will get

  sum - minNum * n = m

This is just a math calculation.

C++代码（Visual Studio 2017）:

#include "stdafx.h"
#include <iostream>
#include <vector>
#include<algorithm> 
using namespace std;

class Solution {
public:
	int minMoves(vector<int>& nums) {
		int res=0;
		int sum = 0;
		int m = *min_element(nums.begin(),nums.end());
		for (int i = 0; i < nums.size(); i++)
			sum += nums[i];
		//cout << sum;
		return sum- m*nums.size();
	}
};

int main()
{
	Solution s;
	vector<int> nums = { 1,2,3 };
	int result;
	result = s.minMoves(nums);
	cout << result;
    return 0;
}