373. Find K Pairs with Smallest Sums

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You are given two integer arrays nums1 and nums2 sorted in ascending order and an integer k.

Define a pair $(u,v)$ which consists of one element from the first array and one element from the second array.

Find the $k$ pairs $(u_1,v_1),(u_2,v_2) ...(u_k,v_k)$ with the smallest sums.

Example 1:

Input: nums1 = [1,7,11], nums2 = [2,4,6], k = 3
Output: [[1,2],[1,4],[1,6]] 
Explanation: The first 3 pairs are returned from the sequence: 
             [1,2],[1,4],[1,6],[7,2],[7,4],[11,2],[7,6],[11,4],[11,6]

Example 2:

Input: nums1 = [1,1,2], nums2 = [1,2,3], k = 2
Output: [1,1],[1,1]
Explanation: The first 2 pairs are returned from the sequence: 
             [1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3]

Example 3:

Input: nums1 = [1,2], nums2 = [3], k = 3
Output: [1,3],[2,3]
Explanation: All possible pairs are returned from the sequence: [1,3],[2,3]

分析

寻找和最小的k对数字。最直接的方法是给出所有的对，一共有$n_1 \times n_2$对数字，其中$n_1$和$n_2$分别为数组nums1和nums2的长度。然后将所有对排序，取出前$k$对数字。那么怎么排序呢？肯定是必须构造一个Comparator。

public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
    List<int[]> list = new ArrayList<>();
    if (nums1 == null || nums2 == null) return list;
    int n1 = nums1.length, n2 = nums2.length;
    PriorityQueue<int[]> pq = new PriorityQueue<>(new PairComparator());
    for (int i = 0; i < n1; i++)
        for (int j = 0; j < n2; j++)
            pq.offer(new int[]{nums1[i], nums2[j]});

    int bound = Math.min(k, n1*n2);
    for (int i = 0; i < bound; i++)
        list.add(pq.poll());
    return list;
}


class PairComparator implements Comparator<int[]> {
    public int compare(int[] one, int[] two) {
        return one[0]  + one[1] - two[0] - two[1];
    }
}

一种更好的方法是使用二叉堆来保存最小对，但是只需要维护K个对即可。因为对于每一个在nums1中的元素来说，对于最小的对，总是从nums2中的第一个元素开始，因为nums2是排序的。所以下一个元素总是nums1[this specific number] + nums2[current_associated_index + 1]，除非越界。

public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
    PriorityQueue<int[]> pq = new PriorityQueue<>((a,b) -> a[0]+a[1]-b[0]-b[1]);
    List<int[]> list = new ArrayList<>();
    if (nums1 == null || nums2 == null) return list;
    int n1 = nums1.length, n2 = nums2.length;
    if (n1 == 0 || n2 == 0 || k == 0) return list;
    for (int i = 0; i < n1 && i < k; i++) 
        // nums1中的值，nums2中的值， nums2的下标
        pq.offer(new int[]{nums1[i], nums2[0], 0});
    int[] cur;
    int bound = Math.min(n1*n2, k);
    while (k--> 0 && !pq.isEmpty()){
        cur = pq.poll();
        list.add(new int[]{cur[0], cur[1]});
        if (cur[2] + 1 == n2) continue;
        pq.offer(new int[]{cur[0], nums2[cur[2]+1], cur[2]+1});
    }
    return list;
}