973. K Closest Points to Origin

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LeetCode

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Problem

Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane and an integer k, return the k closest points to the origin (0, 0).

The distance between two points on the X-Y plane is the Euclidean distance (i.e., √(x1 - x2)2 + (y1 - y2)2).

You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in).

Example 1:

Input: points = [[1,3],[-2,2]], k = 1
Output: [[-2,2]]
Explanation:
The distance between (1, 3) and the origin is sqrt(10).
The distance between (-2, 2) and the origin is sqrt(8).
Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin.
We only want the closest k = 1 points from the origin, so the answer is just [[-2,2]].

Example 2:

Input: points = [[3,3],[5,-1],[-2,4]], k = 2
Output: [[3,3],[-2,4]]
Explanation: The answer [[-2,4],[3,3]] would also be accepted.

Constraints:

• 1 <= k <= points.length <= 104
• -104 <= xi, yi <= 104

Code

class Solution {
    public int[][] kClosest(int[][] points, int K) {
        int left = 0;
        int right = points.length - 1;
        
        while (left <= right) {
            int mid = helper(points, left, right);
            if (mid == K)
                return Arrays.copyOfRange(points, 0, K);
            if (mid > K) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }

        return points;
    }

    private int helper(int[][] nums, int left, int right) {
        int[] pivot = nums[left];

        int l = left + 1;
        int r = right;

        // 给l正确的位置
        // 右边的数比l大 左边的数比l小
        while (l <= r) {
            if (compare(nums[l], pivot) > 0 && compare(nums[r], pivot) < 0) {
                swap(nums, l, r);
                l++;
                r--;
            }

            // 跳过符合条件的值
            if (compare(nums[l], pivot) <= 0)
                l++;
            if (compare(nums[r], pivot) >= 0)
                r--;
        }

        swap(nums, left, r);

        return r;
    }

    // 比较到0,0的距离
    private int compare(int[] p1, int[] p2) {
        return p1[0] * p1[0] + p1[1] * p1[1] - p2[0] * p2[0] - p2[1] * p2[1];
    }

    private void swap(int[][] nums, int left, int right) {
        int[] tmp = nums[left];
        nums[left] = nums[right];
        nums[right] = tmp;
    }
}