leetcode 349——两个数组的交集

一.java

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

        for(int i:nums2)
        {
            if(set1.contains(i))
            {
                set2.add(i);
            }
        }

        int[] arr = new int[set2.size()];
        int j = 0;

        for(int i:set2)
        {
            arr[j++] = i;
        }

        return arr;
    }
}

二.双指针

int cmp(void* a, void* b) 
{
    return *(int*)a - *(int*)b;
}

int* intersection(int* nums1, int nums1Size, int* nums2, int nums2Size, int* returnSize) {
    qsort(nums1, nums1Size, sizeof(int), cmp);
    qsort(nums2, nums2Size, sizeof(int), cmp);
    *returnSize = 0;
    int index1 = 0, index2 = 0;
    int* intersection = malloc(sizeof(int) * (nums1Size + nums2Size));

    while (index1 < nums1Size && index2 < nums2Size) 
    {
        int num1 = nums1[index1], num2 = nums2[index2];
        if (num1 == num2) 
        {
            // 保证加入元素的唯一性
            if (!(*returnSize) || num1 != intersection[(*returnSize) - 1]) {
                intersection[(*returnSize)++] = num1;
            }

            index1++;
            index2++;
        } 
        else if (num1 < num2) 
        {
            index1++;
        } 
        else 
        {
            index2++;
        }
    }

    return intersection;
}