Code & Func

第20天。

今天的题目是 Minimum Moves to Equal Array Elements II ：

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:

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Input:
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[1,2,3]
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Output:
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2
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Explanation:
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Only two moves are needed (remember each move increments or decrements one element):
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[1,2,3]  =>  [2,2,3]  =>  [2,2,2]

这道题需要一些数学推导，它的目标就是：

$$ min_k { \sum_{i=1}^n |n_i - n_k| } $$ 其中 $n_i$ 表示数组排序后中第 $i$ 个元素。

我们将式子展开可以得到： $$ min_k { \sum_{i=1}^n |n_i - n_k| } =

min_k { \sum_{i=1}^k (n_k-n_i) + \sum_{i=k+1}^n(n_i-n_k) } \

= min_k { \sum_{i=1}^k n_k - \sum_{i=1}^k n_i + \sum_{i=k+1}^n n_i - \sum_{i=k+1}^n n_k } \

= min_k { \sum_{i=k+1}^n n_i - \sum_{i=1}^k n_i + (2k - n)n_k } $$ 因此，我们可以写出如下代码：

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int minMoves2(vector<int>& nums) {
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    long long res = LONG_MAX;
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    sort(nums.begin(), nums.end());
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    long long rightSum = 0;
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    for(auto i: nums) rightSum += i;
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    long long leftSum = 0;
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    int n = nums.size();
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    for(int i = 0;i < n; ++i) {
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        res = min(res, rightSum - leftSum + (2*i - n) * (long long)nums[i]);
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        rightSum -= nums[i];
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        leftSum += nums[i];
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        // cout << res << endl;
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    }
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    return res;
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}

这样还不是最优解，然而最优解我没看懂（捂脸），为什么用中位数求就是对的呢？：

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    int minMoves2(vector<int>& nums) {
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        sort(nums.begin(), nums.end());
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        int mid;
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        if (nums.size() % 2 == 0){
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            mid = (nums[nums.size()/2] + nums[(nums.size()/2) - 1])/2;
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        }else{
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            mid = nums[nums.size()/2];
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        }
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        int result = 0;
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        for (int i = 0; i < nums.size(); i++){
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            result += abs(nums[i] - mid);
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        }
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        return result;
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    }