题目
求两数组交集
Given two arrays, write a function to compute their intersection.
Example 1:
Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]
Example 2:
Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]
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?
思路
值得注意的是，题目中说结果可以是无序的，这意味着算法执行过程中可以改变原数组的顺序，自然就想到通过排序解决：先将两个数组进行排序，然后使用双指针法依次比对两个数组——
- 当各自指针指向的元素相等时，保存下来（因为长度是动态的，所以使用ArrayList这个集合来保存），并将两个指针同时往后移动
- 当元素不相等时，只需要移动较小元素的指针即可
排序操作决定了时间复杂度：O(nlgn)
代码
class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
//先对两个数组进行排序
Arrays.sort(nums1);
Arrays.sort(nums2);
ArrayList result = new ArrayList();//保存交集中的uansu
for(int i = 0, j = 0; i < nums1.length && j < nums2.length ;){
if(nums1[i] == nums2[j]){//元素相等
result.add(nums1[i]);//保存
i++;//同时移动索引进行后面的比较
j++;
}
else if(nums1[i] < nums2[j])
i++;
else
j++;
}
//把 ArrayList中的元素转换为数组返回
int[] res = new int[result.size()];
for(int i = 0; i < res.length; i++){
res[i] = (int)result.get(i);
}
return res;
}
}
Runtime: 1 ms, faster than 100.00% of Java online submissions for Intersection of Two Arrays II.
Memory Usage: 35 MB, less than 74.60% of Java online submissions for Intersection of Two Arrays II.
