453. Minimum Moves to Equal Array Elements

原创于 2018-04-02 17:29:43 发布 · 102 阅读

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本文探讨了一种算法，用于确定将非空整数数组的所有元素变为相等所需的最小步数，每一步可以将n-1个元素递增1。通过求解数学方程组，该文提供了一个简洁高效的解决方案。

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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]

数学问题，没做过的话完全没办法啊，，，

摘自讨论区：@wang.senyuan

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.

class Solution {
public:
    int minMoves(vector<int>& nums) {
        int sum=0,minNum=nums[0];
        for(auto i:nums)
        {
            sum +=i;
            minNum=min(minNum,i);
        }
        int m=sum-minNum*nums.size();
        return m;    
    }
};