658. 找到 K 个最接近的元素

题目：

思路：

【1】这道题的核心在于找到X的值或者距离X值之间差值最小的值，然后往两边进行扩散，差值最小的优先纳入。

代码展示：

//时间3 ms 击败99.5%
//内存43.8 MB 击败 79.45%
class Solution {
    public List<Integer> findClosestElements(int[] arr, int k, int x) {
        int right = binarySearch(arr, x);
        int left = right - 1;
        while (k-- > 0) {
            if (left < 0) {
                right++;
            } else if (right >= arr.length) {
                left--;
            } else if (x - arr[left] <= arr[right] - x) {
                left--;
            } else {
                right++;
            }
        }
        List<Integer> ans = new ArrayList<Integer>();
        for (int i = left + 1; i < right; i++) {
            ans.add(arr[i]);
        }
        return ans;
    }

    public int binarySearch(int[] arr, int x) {
        int low = 0, high = arr.length - 1;
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] >= x) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }
        return low;
    }
}

//时间0 ms 击败 100%
//内存44 MB 击败 42.69%
import java.util.AbstractList;
class Solution {
    public List<Integer> findClosestElements(int[] arr, int k, int x) {
        int left=0;
        int right=arr.length-k;
        List<Integer> list = new ArrayList<>();
        while(left<right){
            int temp=left+(right-left)/2;
            if(x - arr[temp] > arr[temp + k] - x){
                left=temp+1;
            }else {
                right=temp;
            }
                
        }
        int finalI = left;
        return new AbstractList<Integer>() {
            @Override
            public Integer get(int index) {
                return arr[index+ finalI];
            }

            @Override
            public int size() {
                return k;
            }
        };   
    }
}