349. Intersection of Two Arrays

求两个set的交集，注意集合类和数组之间的转换

1.set做法

public int[] intersection(int[] nums1, int[] nums2) {
    // add nums1 to set
    Set<Integer> set1 = new HashSet<>();
    for (int i : nums1) set1.add(i);

    // create result set, check if nums2 is in set1
    Set<Integer> set = new HashSet<>();
    for (int i : nums2) {
        if (set1.contains(i)) set.add(i);
    }

    // create result int[]
    int[] res = new int[set.size()];
    int j = 0;
    for (int i : set) {
        res[j++] = i;
    }
    return res;
}

2.参考350，双指针

public int[] intersection(int[] nums1, int[] nums2) {
    // boundary check here
    if (nums1 == null || nums2 == null) return new int[]{};

    // create result set
    Set<Integer> set = new HashSet<>();

    // sort
    Arrays.sort(nums1);
    Arrays.sort(nums2);

    // two pointers
    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 {
            set.add(nums1[i]);
            i++;
            j++;
        }
    }

    // copy set to res[]
    int[] res = new int[set.size()];
    int k = 0;
    for (int num : set) {
        res[k++] = num;
    }
    return res;
}

3.调api，求集合交集，慢

return list(set(nums1) & set(nums2))

public int[] intersection(int[] nums1, int[] nums2) {
    Set<Integer> set1 = new HashSet<Integer>();
    Set<Integer> set2 = new HashSet<Integer>();
    for (int num : nums1) set1.add(num);
    for (int num : nums2) set2.add(num);

    set1.retainAll(set2);

    int i = 0;
    int[] res = new int[set1.size()];
    for (int num : set1) {
        res[i++] = num;
    }

    return res;
}

retainAll的实现方式，可以看到比contains多了些代码:

public boolean retainAll(Collection<?> c) {
    Objects.requireNonNull(c);
    boolean modified = false;
    Iterator<E> it = iterator();
    while (it.hasNext()) {
        if (!c.contains(it.next())) {
            it.remove();
            modified = true;
        }
    }
    return modified;
}