2022.03.18(LC_349_两个数组的交集)

方法一：哈希表

--单Set

class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        Set<Integer> set = new HashSet<>();
        for (int num : nums1) {
            set.add(num);
        }
        int size = 0;
        for (int num : nums2) {
            if (set.contains(num)) {
                nums1[size++] = num;
                set.remove(num);
            }
        }
        int[] res = new int[size];
        for (int i = 0; i < size; i++) {
            res[i] = nums1[i];
        }
        return res;
    }
}

--双Set

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

方法二：排序+双指针

class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        Set<Integer> set = new HashSet<>();
        int i = 0, j = 0;
        while (i < nums1.length && j < nums2.length) {
            if (nums1[i] == nums2[j]) {
                set.add(nums1[i]);
                i++;
                j++;
            } else if (nums1[i] < nums2[j]) {
                i++;
            } else {
                j++;
            }
        }
        int k = 0;
        int[] res = new int[set.size()];
        for (int num : set) {
            res[k++] = num;
        }
        return res;
    }
}