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# 題目敘述
Given two 0-indexed integer arrays nums1
and nums2
, return a list answer
of size 2
where:
answer[0]
is a list of all distinct integers innums1
which are not present innums2
.answer[1]
is a list of all distinct integers innums2
which are not present innums1
.
Note that the integers in the lists may be returned in any order.
# Example 1
Input: nums1 = [1,2,3], nums2 = [2,4,6]
Output: [[1,3],[4,6]]
Explanation:
For nums1, nums1[1] = 2 is present at index 0 of nums2, whereas nums1[0] = 1 and nums1[2] = 3 are not present in nums2. Therefore, answer[0] = [1,3].
For nums2, nums2[0] = 2 is present at index 1 of nums1, whereas nums2[1] = 4 and nums2[2] = 6 are not present in nums2. Therefore, answer[1] = [4,6].
# Example 2
Input: nums1 = [1,2,3,3], nums2 = [1,1,2,2]
Output: [[3],[]]
Explanation:
For nums1, nums1[2] and nums1[3] are not present in nums2. Since nums1[2] == nums1[3], their value is only included once and answer[0] = [3].
Every integer in nums2 is present in nums1. Therefore, answer[1] = [].
class Solution { | |
public: | |
vector<vector<int>> findDifference(vector<int>& nums1, vector<int>& nums2) { | |
vector<vector<int>> res; | |
vector<int> tmp; | |
map<int, int> mp1, mp2; | |
for (int n : nums1) mp1[n] = 1; | |
for (int n : nums2) mp2[n] = 1; | |
for (auto m : mp1) { | |
if (mp2.count(m.first) == 0) tmp.push_back(m.first); | |
} | |
res.push_back(tmp); | |
tmp.clear(); | |
for (auto m : mp2) { | |
if (mp1.count(m.first) == 0) tmp.push_back(m.first); | |
} | |
res.push_back(tmp); | |
return res; | |
} | |
}; |
import java.util.ArrayList; | |
import java.util.List; | |
class Solution { | |
public List<List<Integer>> findDifference(int[] nums1, int[] nums2) { | |
List<List<Integer>> ans = new ArrayList<>(); | |
List<Integer> map1 = new ArrayList<>(), map2 = new ArrayList<>(); | |
for(int n : nums1) map1.add(n); | |
for(int n : nums2) map2.add(n); | |
List<Integer> temp1 = new ArrayList<>(); | |
for(int n : nums1){ | |
if(!map2.contains(n) && !temp1.contains(n)){ | |
temp1.add(n); | |
} | |
} | |
ans.add(temp1); | |
List<Integer> temp2 = new ArrayList<>(); | |
for(int n : nums2){ | |
if(!map1.contains(n) && !temp2.contains(n)){ | |
temp2.add(n); | |
} | |
} | |
ans.add(temp2); | |
return ans; | |
} | |
} |
class Solution: | |
def findDifference(self, nums1, nums2): | |
return [list(set(nums1) - set(nums2)), list(set(nums2) - set(nums1))] |