leetcode note--leetcode 350 Intersection of Two Arrays II

350. Intersection of Two Arrays II

 

Add to List

QuestionEditorial Solution

 My Submissions

• Total Accepted: 43789
• Total Submissions: 101360
• Difficulty: Easy
• Contributors: Admin

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

Note:

• Each element in the result should appear as many times as it shows in both arrays.
• The result can be in any order.

Follow up:

• What if the given array is already sorted? How would you optimize your algorithm?
• What if nums1's size is small compared to nums2's size? Which algorithm is better?

• What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

个人觉得应该先做这个，再做349题，这个比那个少两行。

public class Solution {      public int[] intersect(int[] nums1, int[] nums2) {          List<Integer> list = new ArrayList<>();           Arrays.sort(nums1);          Arrays.sort(nums2);          int len1 = nums1.length;          int len2 = nums2.length;          if(len1==0 || len2==0) return new int[0];          int p1 = 0,p2 = 0;          while(p1<len1 && p2<len2){             /* while(p1<len1-1 && nums1[p1]==nums1[p1+1])p1++;              while(p2<len2-1 && nums2[p2]==nums2[p2+1])p2++;  */            if(nums1[p1]==nums2[p2]){                  list.add(nums1[p1]);                  p1++;p2++;              }else if(nums1[p1]<nums2[p2]){                  p1++;              }else{                  p2++;              }          }          int res[] = new int[list.size()];          for(int i=0;i<list.size();i++){              res[i] = list.get(i);          }          return res;      }  }