LeetCode 658. Find K Closest Elements C++

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

Approach

1. 给你一组数组，K和X，问你离X的差值最接近的K个是哪些，当两差值相同的时候值小的优先，但差值不同时，差值小的优先。这道题很容易想到用二分查找，查找到离X最近小于或等于的值的下标left，接着就要处理差值最近的K个是哪些值，此时这里我们可以用two pointers的思想，从0~left+K范围，缩小到K个，就是答案。当然left+K要注意越界，我们对边界可以有特殊的处理，left=0时，答案即是前K个元素和left=数组的长度-1，答案即是后K个元素，这让会更快。

Code

class Solution {
public:
    vector<int> findClosestElements(vector<int>& arr, int k, int x) {
        if (arr.size() < 2)return arr;
        int left = 0, right = arr.size() - 1, len = arr.size();
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (arr[mid] == x) {
                left = mid;
                break;
            }
            if (arr[mid] > x)right = mid - 1;
            else left = mid + 1;
        }
        if (left == 0) {
            return vector<int>(arr.begin(), arr.begin() + k);
        }
        else if (left == arr.size() - 1 ) {
            return vector<int>(arr.end() - k, arr.end());
        }   
        int i = 0, j = left + k1;
        while (j - i +1 > k) {
            int rr = arr[j] - x;
            int ll = x - arr[i];
            if (rr >= ll)j--;
            else i++;
        }
        return vector<int>(arr.begin() + i, arr.begin() + j+1);
    }
};