[LintCode/LeetCode] Intersection of Two Arrays I &

enda 发布于2019-08-14 14:03 / 3510人阅读

摘要：先想到的是，其实也可以，只是需要在遍历的时候，添加到数组中的数要掉，略微麻烦了一点。在里跑的时候，也要快一点。另一种类似做法的就快的多了。如果是找出所有包括重复的截距呢

Problem

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

Notice

Each element in the result must be unique.
The result can be in any order.

Example

Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].

Challenge

Can you implement it in three different algorithms?

Note

先想到的是HashSet()，其实HashMap也可以，只是需要在遍历nums2的时候，添加到res数组中的数要remove掉，略微麻烦了一点。在LC里跑的时候，HashSet也要快一点。
另一种类似HashMap做法的BitSet()就快的多了。

Solution HashSet

public class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        Set set1 = new HashSet();
        Set set2 = new HashSet();
        List ans = new ArrayList();
        for (int i = 0; i < nums1.length; i++) set1.add(nums1[i]);
        for (int i = 0; i < nums2.length; i++) {
            if (set1.contains(nums2[i])) set2.add(nums2[i]);
        }
        int[] res = new int[set2.size()];
        int index = 0;
        for (Integer i: set2) res[index++] = i;
        return res;
    }
}

BitSet

public class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        int[] res = new int[nums1.length];
        if (nums1.length == 0 || nums2.length == 0) return new int[0];
        int index = 0;
        BitSet set = new BitSet();
        for (int i = 0; i < nums1.length; i++) {
            set.set(nums1[i]);
        }
        for (int i = 0; i < nums2.length; i++) {
            if (set.get(nums2[i]) == true) {
                res[index++] = nums2[i];
                set.set(nums2[i], false);
            }
        }
        return Arrays.copyOfRange(res, 0, index);
    }
}

Follow Up

如果是找出所有包括重复的截距呢？-- Intersection of Two Arrays II

public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        int k = 0, l1 = nums1.length, l2 = nums2.length;
        int[] result = new int[l1];
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        int i = 0, j = 0;
        while (i < l1 && j < l2)
            if (nums1[i] < nums2[j]) i++;
            else if (nums1[i] == nums2[j++]) result[k++] = nums1[i++];
        return Arrays.copyOf(result, k);
    }
}

GPU云服务器云服务器 Intersection Arrays is_subclass_of函数 Arrays类

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