leetcode350. Intersection of Two Arrays II

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给定两个数组，编写一个函数找出它们的交集。例如，给定nums1=[1,2,2,1]，nums2=[2,2]，返回[2,2]。注意：结果中的每个元素应与其在两个数组中出现的次数相同，顺序不限。后续问题包括：已排序数组如何优化算法？当nums1的大小远小于nums2时，哪种算法更优？如果nums2的元素存储在磁盘上，内存有限，无法一次性加载所有元素，怎么办？" 125802457,11412032,互亿无线短信插件在Whatsns_V6.03的安装指南,"['前端开发', 'javascript', 'mybatis']

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350. 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 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?

解法

public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        if (nums1 == null || nums1.length == 0) {
            return nums1;
        }
        if (nums2 == null || nums2.length == 0) {
            return nums2;
        }
        List<Integer> list = new ArrayList<Integer>();
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        int i = 0;
        int j = 0;
        while (i < nums1.length && j < nums2.length) {
            if (nums1[i] < nums2[j]) {
                i++;
            } else if (nums1[i] == nums2[j]) {
                list.add(nums1[i]);
                i++;
                j++;
            } else {
                j++;
            }
        }
        int size = list.size();
        int[] ret = new int[size];
        int k = 0;
        Iterator<Integer> it = list.iterator();
        while (it.hasNext()) {
            ret[k++] = it.next();
        }

        return ret;
    }
}

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