LeetCode 350. Intersection of Two Arrays II [Easy]

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题目来源：https://leetcode.com/problems/intersection-of-two-arrays-ii

题目难度：Easy

题目

Given two arrays, write a function to compute their intersection.

Example 1:

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

Example 2:

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Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]

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?

解答1[Java]

思路

借助一个 Map，先统计数组1中每一个数组的个数，然后再遍历数组2，如果 Map 中存在此元素，则加入到结果数组，并且把 Map 中统计的数量减去 1。

实现

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public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        HashMap<Integer, Integer> map = new HashMap<>();
        ArrayList<Integer> result = new ArrayList<>();
        for (int value : nums1) {
            if (map.containsKey(value)) {
                map.put(value, map.get(value) + 1);
            } else {
                map.put(value, 1);
            }
        }

        for (int value : nums2) {
            if (map.containsKey(value) && map.get(value) > 0) {
                result.add(value);
                map.put(value, map.get(value) - 1);
            }
        }

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

        return resultArray;
    }
}

复杂度分析

HashMap 的 get() 时间复杂度为 $O(1)$。

解答2[Java]

思路

先对两个数组进行排序，然后两个数组分别设一个指针，判断当前指针指向的元素是否相等，如果相等就把指针当前指向的元素加入结果数组。如果不相等，哪个指向的元素更小，哪个指针向右移动，另外一个指针保持位置不变。

实现

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class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        List<Integer> intersection = new ArrayList<>();
        Arrays.sort(nums1);
        Arrays.sort(nums2);

        int i = 0;
        int j = 0;
        while (i < nums1.length && j < nums2.length) {
            if (nums1[i] == nums2[j]) {
                intersection.add(nums2[j]);
                i++;
                j++;
            } else if (nums1[i] > nums2[j]) {
                j++;
            } else {
                // nums[i] < nums[j]
                i++;
            }
        }


        int[] res = new int[intersection.size()];
        for (int k = 0; k < intersection.size(); k++) {
            res[k] = intersection.get(k);
        }
        
        return res;
    }
}