[leetcode-658-Find K Closest Elements]

Given a sorted array, two integers k and x, find the k closest elements to x in the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred.

Example 1:

Input: [1,2,3,4,5], k=4, x=3
Output: [1,2,3,4]

Example 2:

Input: [1,2,3,4,5], k=4, x=-1
Output: [1,2,3,4]

Note:

1. The value k is positive and will always be smaller than the length of the sorted array.
2. Length of the given array is positive and will not exceed 10⁴
3. Absolute value of elements in the array and x will not exceed 10⁴

思路：

用一个map记录数组中的值距离x的大小，利用map有序的特性。

        int SIZE = arr.size();
        map<int, vector<int>> m;
        for(int i = ; i < SIZE; ++i) {
            int val = arr[i];
            m[abs(val - x)].push_back(val);
        }
        vector<int> ans;
        auto it = m.begin();
        while(ans.size() < k) {
            vector<int> &v = it->second;
            sort(v.begin(), v.end());
            int i = ;
            while(i < v.size() && k > ans.size()) {
                ans.push_back(v[i++]);
            }
            ++it;
        }
        sort(ans.begin(), ans.end());
        return ans;

vector<int> findClosestElements(vector<int>& arr, int k, int x)
{
    vector< int > ret;
    vector< int > cur;
    auto it = lower_bound( arr.begin(), arr.end(), x );//低
    //cout << *it << endl;

    long long sum = ;
    long long min_val = 0xc0c0c0c0;
    auto it_start = ( it - k < arr.begin() ) ? arr.begin() : it-k;
    auto it_end = ( it > arr.end() - k ) ? arr.end() - k : it;

    //cout << *it_start << endl;
    //cout << *it_end << endl;

    for( auto it_cur = it_start; it_cur <= it_end; it_cur++ )
    {
        sum = ;
        cur.clear();
        for( int i = ; i < k; i++ )
        {
            cur.push_back( *(it_cur+i) );
            sum += abs( ( *(it_cur+i) - x ) );
        }
        if( sum < min_val )
        {
            min_val = sum;
            swap( ret, cur );
        }
    }

    return ret;
}