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?
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題目

計算兩個數組的交

樣例
nums1 = [1, 2, 2, 1], nums2 = [2, 2], 返回 [2, 2].

分析

這道題是上一道題的擴展，只是這次要記錄重復的元素的個數，這次我們就用一個哈希表，鍵記錄重復元素，值記錄重復個數就行了。采用hashset的方法

代碼

public class Solution {
    /**
     * @param nums1 an integer array
     * @param nums2 an integer array
     * @return an integer array
     */
    public int[] intersection(int[] nums1, int[] nums2) {
        // Write your code here
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for(int i = 0; i < nums1.length; ++i) {
            if (map.containsKey(nums1[i]))
                map.put(nums1[i], map.get(nums1[i]) + 1); 
            else
                map.put(nums1[i], 1);
        }

        List<Integer> results = new ArrayList<Integer>();

        for (int i = 0; i < nums2.length; ++i)
            if (map.containsKey(nums2[i]) &&
                map.get(nums2[i]) > 0) {
                results.add(nums2[i]);
                map.put(nums2[i], map.get(nums2[i]) - 1); 
            }

        int result[] = new int[results.size()];
        for(int i = 0; i < results.size(); ++i)
            result[i] = results.get(i);

        return result;
    }
}