[LeetCode] 350. Intersection of Two Arrays II

Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.

Example 1:

Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]

Example 2:

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]
Explanation: [9,4] is also accepted.

Constraints:

• 1 <= nums1.length, nums2.length <= 1000
• 0 <= nums1[i], nums2[i] <= 1000

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?

两个数组的交集II。

题意跟349题几乎一样，唯一的区别在于349题请输出unique的交集，但是本题请输出所有的交集，所以需要考虑重复数字的情况。

两种做法，一是hashmap，二是双指针。

hashmap的做法是先扫描num1，记录每个数字和他们出现的次数；再遍历num2，如果在hashmap里出现了相同的key则value--，但是注意判断的方式不是if (map.containsKey(x))，而是判断这个key背后的value是否大于0，因为之后要对value--。

时间O(n)

空间O(n)

Java实现

class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        HashMap<Integer, Integer> map = new HashMap<>();
        List<Integer> list = new ArrayList<>();
        for (int num : nums1) {
            if (map.containsKey(num)) {
                map.put(num, map.get(num) + 1);
            } else {
                map.put(num, 1);
            }
        }
        for (int num : nums2) {
            if (map.containsKey(num)) {
                if (map.get(num) > 0) {
                    list.add(num);
                    map.put(num, map.get(num) - 1);
                }
            }
        }
        int[] res = new int[list.size()];
        int k = 0;
        for (int num : list) {
            res[k++] = num;
        }
        return res;
    }
}

JavaScript实现

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersect = function(nums1, nums2) {
    let map = new Map();
    let list = [];
    for (let num of nums1) {
        if (map.has(num)) {
            map.set(num, map.get(num) + 1);
        } else {
            map.set(num, 1);
        }
    }
    for (let num of nums2) {
        if (map.get(num) > 0) {
            list.push(num);
            map.set(num, map.get(num) - 1);
        }
    }
    return list;
};

双指针的做法是先sort两个数组然后逐个比较，看两边遍历到的元素是否一样，若是则加入结果集。

时间O(nlogn) - sort

空间O(n)

Java实现ONLY

class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        List<Integer> list = new ArrayList<>();
        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]) {
                j++;
            } else {
                list.add(nums1[i]);
                i++;
                j++;
            }
        }
        int[] res = new int[list.size()];
        int k = 0;
        for (int num : list) {
            res[k++] = num;
        }
        return res;
    }
}

LeetCode 题目总结