462. Minimum Moves to Equal Arra

这道题一直在思考O(n)解法浪费了很多时间，应当保证能做出的情况下去思考比较好。。。自己写的效率还行。

我的解法

class Solution {
public:
    int minMoves2(vector<int>& nums) {
        int output = 0;
        sort(nums.begin(), nums.end());
        int middle = nums[nums.size()/2];
        for (int i = 0; i < nums.size(); i ++)
            output += abs(nums[i] - middle);
        return output;
    }
};

别人的解法

• 别人的解法1

这里这个nth_element时间复杂度只有O(n)，很厉害的样子。

// O(n).
// Imagine the nums are sorted, and the final value is k, we start find k from the first element.
// If we increase k, the elements <= k will need move one step more, and the elements > k will need to move one step less.
// If there are more elements > k than elements <= k, we should increase k to minimize the moves.
// So we just increase k, until k reach the median of of the nums array. By then, the number of elements <= k equals to that of elements > k.
// (There is a slight different when the number of array is odd, but it's similar).
// If we keep increasing k after k reach the median of the array, more numbers >k than <= k, and more moves needed, so we should stop.
//
// The sort is not needed since we find the k is the median of the array, there is an average O(n) algorithm to find such k.
class Solution {
public:
    int minMoves2(vector<int>& nums) {
        int n = nums.size();
        auto it = nums.begin() + n/2;
        nth_element(nums.begin(), it, nums.end());
        int median = *it;
        int total = 0;
        for (auto &i : nums)
            total += abs(i-median);
        return total;
    }
};

• 别人的解法2

思考结果本质上是一样的，但是思维的过程不太一样，可以想一想。

int minMoves2(vector<int>& nums) {
      sort(nums.begin(), nums.end()); int n = nums.size(), res = 0;
      for (int i = 0; i < n/2; ++i) res += (nums[n-1-i]-nums[i]);
      return res;
    }