Minimum Moves to Equal Array Elements问题及解法
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问题描述:
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.
示例:
Input:[1,2,3]Output:3Explanation:Only three moves are needed (remember each move increments two elements):[1,2,3] => [2,3,3] => [3,4,3] => [4,4,4]问题分析:
我们做如下定义:
sum:未做任何移动前的所有元素之和
minNum:未做任何移动前的最小元素值
n:是数组的大小
m:最少移动的次数
x:最少移动次数完成后,每个元素的大小
因此我们有等式sum + m * (n -1) = x * n,x = minNum + m 成立,也就是sum - minNum * n = m。
过程详见代码:
class Solution {public: int minMoves(vector<int>& nums) { int mi = nums[0] ,sum = mi; for(int i = 1; i < nums.size(); i++) {sum += nums[i]; mi = min(nums[i],mi);} return sum - nums.size() * mi; }};
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