[LeetCode] 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

题目：题目要求找出排序数组中距离x的k个数，如果x左右两边的值距离相同，则保留x左边的数字。

思路：通过题意，也就是求一个长度为k的子数组，要求数组中的值都距离x最近，则可以从数组两边开始删除，因为距离最远的元素肯定出现在首尾。比较数组的首尾元素与x的距离，删除距离较大的那个元素。这样当数组中元素剩余k个时，就是题目要求的结果数组。代码如下

class Solution {
public:
    vector<int> findClosestElements(vector<int>& arr, int k, int x) {
        vector<int> res = arr;
        while (res.size() > k)
        {
            if (x - res.front() <= res.back() - x)
                res.pop_back();
            else
                res.erase(res.begin());
        }
        return res;
    }
};