453. Minimum Moves to Equal Array Elements*

453. Minimum Moves to Equal Array Elements*

https://leetcode.com/problems/minimum-moves-to-equal-array-elements/

题目描述

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]

C++ 实现 1

参考: 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.

也就是说, 必须注意到, 数组中的最小值 minNum 在经过 m 次变化之后的值为 minNum + m, 并且等于其他值 x = minNum + m, 此时总和一方面可以通过初始的 sum 加上 m * (n - 1) (每次都是剩下的 n - 1 个元素加 1) 计算, 或者通过最终数组中所有值都相等进行计算 n * x, 那么可以得到公式 m = sum - minNum * n.

下面代码中 sum 设置为 long long 是为了避免 [1, INT32_MAX] 这种极端情况, 注意求和的时候直接使用

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std::accumulate(nums.begin(), nums.end(), 0)

是会报错的.

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

C++ 实现 2

或者这样也行, 避免数值过大的极端情况.

class Solution {
public:
    int minMoves(vector<int>& nums) {
        int res=0,minelement=*min_element(nums.begin(), nums.end());
        for(int i:nums)res+=i-minelement;
        return res;
    }
};