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  • LeetCode 373. Find K Pairs with Smallest Sums

    原题链接在这里:https://leetcode.com/problems/find-k-pairs-with-smallest-sums/

    题目:

    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 (u1,v1),(u2,v2) ...(uk,vk) with the smallest sums.

    Example 1:

    Given nums1 = [1,7,11], nums2 = [2,4,6],  k = 3
    
    Return: [1,2],[1,4],[1,6]
    
    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:

    Given nums1 = [1,1,2], nums2 = [1,2,3],  k = 2
    
    Return: [1,1],[1,1]
    
    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:

    Given nums1 = [1,2], nums2 = [3],  k = 3 
    
    Return: [1,3],[2,3]
    
    All possible pairs are returned from the sequence:
    [1,3],[2,3]

    题解:

    用nums2当row, nums1当列, nums1的每一个元素依次与nums2中的元素相加 就形成了一个向右向下ascending的matrix.

    所以也用一个minHeap来存储第一行, 然后poll k次出来加到res中, 每次poll出来的元素下面的元素加到minHeap中.

    Time Complexity: O(n + klogn), n = min(nums1.length, nums2.length). 选长度较小的array来做row.

    Space: O(n), minHeap size.

    AC Java:

     1 class Solution {
     2     public List<List<Integer>> kSmallestPairs(int[] nums1, int[] nums2, int k) {
     3         List<List<Integer>> res = new ArrayList<>();
     4         if(nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0){
     5             return res;
     6         }
     7         
     8         int m = nums1.length;
     9         int n = nums2.length;
    10         PriorityQueue<int []> minHeap = new PriorityQueue<>((a, b) -> a[2]-b[2]);
    11         for(int j = 0; j<n; j++){
    12             minHeap.add(new int[]{0, j, nums1[0]+nums2[j]});
    13         }
    14         
    15         while(k-->0 && !minHeap.isEmpty()){
    16             int [] cur = minHeap.poll();
    17             res.add(Arrays.asList(nums1[cur[0]], nums2[cur[1]]));
    18             if(cur[0] == m-1){
    19                 continue;
    20             }
    21             
    22             minHeap.add(new int[]{cur[0]+1, cur[1], nums1[cur[0]+1] + nums2[cur[1]]});
    23         }
    24         
    25         return res;
    26     }
    27 }

    类似Kth Smallest Element in a Sorted Matrix.  

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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/6361324.html
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