leetcode note--leetcode 350 Intersection of Two Arrays II-优快云博客

本文介绍了一个计算两个数组交集的算法实现，通过排序并遍历两个数组来找出重复元素，并讨论了不同情况下的优化策略。

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350. Intersection of Two Arrays II

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;  
    }  
}