LeetCode:350,Intersection of Two Arrays II(数组的的相交)

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]

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Example 2：

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

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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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  根据这个题目比较简单，有多种解法：

1.利用map的原理：

 思路先将一个数组存储到map当中，该元素作为key关键字，元素出现的次数作为它的value值。

 private int[] intersect1(int[] arr1, int[] arr2) {
        if (arr1.length == 0 || arr2.length == 0) {
            return new int[]{};
        }
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int num1 : arr1) {
            if (!map.keySet().contains(num1)) {
                map.put(num1, 1);
            } else {
                map.put(num1, map.get(num1) + 1);
            }
        }
        ArrayList<Integer> list = new ArrayList<>();
        for (int num2 : arr2) {
            if (map.keySet().contains(num2) && map.get(num2) > 0) {
                list.add(num2);
                map.put(num2, map.get(num2) - 1);
            }
        }

        int[] newArr = new int[list.size()];
        for (int index = 0; index < list.size(); index++) {
            newArr[index] = list.get(index);
        }
        return newArr;
    }

时间复杂度：O(n);

空间复杂度：O(n);

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2.直接利用指针的方式实现:其中在使用指正有两种方式

2.1,利用压缩数组，把数组中不相等的数去掉

private int[] intersect3(int[] nums1, int[] nums2) {
        Arrays.sort(nums1);
        Arrays.sort(nums2);

        int i = 0;
        int j = 0;
        int count = 0;
        int flag = 0;
        while (i < nums1.length - flag && j < nums2.length) {
            if (nums1[i] < nums2[j]) {
                for (int k = i; k < nums1.length - 1; k++) {
                    nums1[k] = nums1[k + 1];
                }
                flag++;
            } else if (nums1[i] > nums2[j]) {
                j++;
            } else {
                count++;
                i++;
                j++;
            }
        }
        return Arrays.copyOfRange(nums1, 0, count);
    }

时间复杂度：O(n^2)

空间复杂度：O(1)

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2.2 利用复制法，把相等的元素复制到数组头部，最后把它全部复制过来；

 private int[] intersect5(int[] nums1, int[] nums2) {
        Arrays.sort(nums1);
        Arrays.sort(nums2);

        int i = 0;
        int j = 0;
        int k = 0;

        while (i < nums1.length && j < nums2.length) {
            if (nums1[i] > nums2[j]) {
                j += 1;
                continue;
            }
            if (nums1[i] < nums2[j]) {
                i += 1;
                continue;
            }
            nums1[k]=nums1[i];
            k += 1;
            j += 1;
            i += 1;
        }
        return Arrays.copyOfRange(nums1, 0, k);
    }

时间复杂度：O(n*logn)

空间复杂度：O(1)