134.Intersection of Two Arrays II

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 num2'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?

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与133类似 ，只不过是这个题目选择的容器是list

public int[] intersect(int[] nums1, int[] nums2) {/*Step1:初始化及临界值操作*/int len1 = nums1.length;int len2 = nums2.length;int result[];ArrayList<Integer> re = new ArrayList<Integer>();if(len1 * len2 <=0){result = new int[0];return result;}/*Step2:把nums1数组放到LinkedList中*/LinkedList list = new LinkedList();for(int i=0;i<len1;i++){list.add(nums1[i]);}/*Step3:遍历nums2，如果在LinkedList中则拿出这个元素*/for(int i=0;i<len2;i++){if(list.contains(nums2[i])){re.add(nums2[i]);list.remove((Integer)nums2[i]);}}/*Step4:整理返回的结果*/int len = re.size();result = new int[len];for(int i=0;i<len;i++){result[i] = re.get(i);}        return result;    }