[Leetcode] 462. Minimum Moves to Equal Array Elements II 解题报告
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题目:
Given a non-empty integer array, find the minimum number of moves required to make all array elements equal, where a move is incrementing a selected element by 1 or decrementing a selected element by 1.
You may assume the array's length is at most 10,000.
Example:
Input:[1,2,3]Output:2Explanation:Only two moves are needed (remember each move increments or decrements one element):[1,2,3] => [2,2,3] => [2,2,2]
思路:
我开始想当然地认为最终需要达到的数值k是整个数组的平均数,但其实应该是整个数组的中位数。为什么呢?
假设整个数组是已经被排好序的,我们设计算法来找出k:如果我们将k增加1,那么所有比k小的数都要多增加1次,所有比k大的数都可以多减小1次。如果比k大的数要多于比k小的数,那么将k增加1显然有助于减小总的移动步数。而当大于k的数的数目等于小于k的数的数目时,我们就达到了最优解,此时k恰好就是数组的中位数。
采用排序显然可以找到中位数,然而却是杀鸡用牛刀。我们可以用partition的思想找到中位数,其平均时间复杂度是O(n)。
代码:
class Solution {public: int minMoves2(vector<int>& nums) { auto it = nums.begin() + nums.size() / 2; // we initialize it to get a better hint nth_element(nums.begin(), it, nums.end()); int median = *it; int total = 0; for (auto num : nums) { total += abs(num - median); } return total; }};
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