[LeetCode][453] Minimum Moves to Equal Array Elements题解

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对于一个序列：

• 设序列的和开始为sum0,结束为sum1
• 序列长度为n
• 序列的最小值为min

对于一个最小值min,每次加法操作都必然作用在min上

• 共作用k次后得到一个数字x
• 此时所有的数字都是x

那么有

k = sum1/n-min
sum1=sum0+(n-1)k
k=sum0-minn

---

源码

/*
 * [453] Minimum Moves to Equal Array Elements
 *
 * https://leetcode.com/problems/minimum-moves-to-equal-array-elements/description/
 *
 * algorithms
 * Easy (48.47%)
 * Total Accepted:    48.5K
 * Total Submissions: 100.1K
 * Testcase Example:  '[1,2,3]'
 *
 * 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]
 *
 *
 */
class Solution {
public:
    int minMoves(vector<int>& nums) {
        int n=nums.size();
        int min=nums[0],sum=0;
        
        for(auto num:nums)
        {
            if(num<min)
            {
                min=num;
            }
            sum+=num;
        }
        return sum-min*n;
    }
};