Given two arrays, write a function to compute their intersection.
Example 1:
Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2]
Example 2:
Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [9,4]
Note:
- Each element in the result must be unique.
- The result can be in any order.
class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
TreeSet<Integer> set = new TreeSet<>(); //集合set自動去重num1
for(int num: nums1)
set.add(num);
ArrayList<Integer> list = new ArrayList<>();
//list列表記錄set包含的Num2元素 並且remove去重set中包含的防止重複添加
for(int num: nums2){
if(set.contains(num)){
list.add(num);
set.remove(num);
}
}
int[] res = new int[list.size()]; //返回數組用來記錄list中的int
for(int i = 0 ; i < list.size() ; i ++)
res[i] = list.get(i);
return res;
}
}
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]
可以利用上面的算法 也可以使用映射
public class solution1 {
public int[] intersect(int[] nums1, int[] nums2) {
TreeMap<Integer,Integer> map=new TreeMap<>();
for(int num:nums1){
if(!map.containsKey(num))
map.put(num,1);
else
map.put(num ,map.get(num)+1);
}
ArrayList<Integer> list=new ArrayList<>();
for(int num:nums2){
if(map.containsKey(num)){
list.add(num);
map.put(num,map.get(num)-1);
if(map.get(num)==0)
map.remove(num);
}
}
int[] res=new int[list.size()];
for (int i=0;i<list.size();i++){
res[i]=list.get(i);
}
return res;
}
}