LeetCode 349. Intersection of Two Arrays
原题链接在这里：https://leetcode.com/problems/intersection-of-two-arrays/
题目：
Given two arrays, write a function to compute their intersection.
Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].
Note:
- Each element in the result must be unique.
- The result can be in any order.
题解：
用一个HashSet 来保存nums1的每个element.
再iterate nums2, 若HashSet contains nums2[i], 把nums2[i]加到res中，并把nums[i]从HashSet中remove掉.
Time Complexity: O(nums1.length + nums2.length). Space: O(nums1.length).
AC Java:
1 public class Solution { 2 public int[] intersection(int[] nums1, int[] nums2) { 3 HashSet<Integer> nums1Hs = new HashSet<Integer>(); 4 for(int num : nums1){ 5 nums1Hs.add(num); 6 } 7 8 List<Integer> res = new ArrayList<Integer>(); 9 for(int num : nums2){ 10 if(nums1Hs.contains(num)){ 11 res.add(num); 12 nums1Hs.remove(num); 13 } 14 } 15 int [] resArr = new int[res.size()]; 16 int i = 0; 17 for(int num : res){ 18 resArr[i++] = num; 19 } 20 return resArr; 21 } 22 }
AC C++:
1 class Solution { 2 public: 3 vector<int> intersection(vector<int>& nums1, vector<int>& nums2) { 4 vector<int> res; 5 unordered_set<int> set(nums1.begin(), nums1.end()); 6 for(int num : nums2){ 7 if(set.erase(num)){ 8 res.push_back(num); 9 } 10 } 11 12 return res; 13 } 14 };
也可以使用两个HashSet.
Time Complexity: O(nums1.length + nums2.length). Space: O(nums1.length).
AC Java:
1 public class Solution { 2 public int[] intersection(int[] nums1, int[] nums2) { 3 HashSet<Integer> nums1Hs = new HashSet<Integer>(); 4 HashSet<Integer> intersectHs = new HashSet<Integer>(); 5 for(int num : nums1){ 6 nums1Hs.add(num); 7 } 8 for(int num : nums2){ 9 if(nums1Hs.contains(num)){ 10 intersectHs.add(num); 11 } 12 } 13 14 int [] res = new int[intersectHs.size()]; 15 int i = 0; 16 for(int num : intersectHs){ 17 res[i++] = num; 18 } 19 return res; 20 } 21 }
Sort nums1 and nums2, 再用双指针 从头iterate两个sorted array.
Time Complexity: O(nlogn). Space: O(1).
AC Java:
1 public class Solution { 2 public int[] intersection(int[] nums1, int[] nums2) { 3 Arrays.sort(nums1); 4 Arrays.sort(nums2); 5 HashSet<Integer> hs = new HashSet<Integer>(); 6 int i = 0; 7 int j = 0; 8 while(i<nums1.length && j<nums2.length){ 9 if(nums1[i] < nums2[j]){ 10 i++; 11 }else if(nums1[i] > nums2[j]){ 12 j++; 13 }else{ 14 hs.add(nums1[i]); 15 i++; 16 j++; 17 } 18 } 19 20 int [] resArr = new int[hs.size()]; 21 int k = 0; 22 for(int num : hs){ 23 resArr[k++] = num; 24 } 25 return resArr; 26 } 27 }
AC C++:
1 class Solution { 2 public: 3 vector<int> intersection(vector<int>& nums1, vector<int>& nums2) { 4 sort(nums1.begin(), nums1.end()); 5 sort(nums2.begin(), nums2.end()); 6 int len1 = nums1.size(); 7 int len2 = nums2.size(); 8 int i = 0; 9 int j = 0; 10 vector<int> res; 11 12 while(i < len1 && j < len2){ 13 if(i > 0 && nums1[i] == nums1[i - 1]){ 14 i++; 15 continue; 16 } 17 18 if(nums1[i] < nums2[j]){ 19 i++; 20 }else if(nums1[i] > nums2[j]){ 21 j++; 22 }else{ 23 res.push_back(nums1[i]); 24 i++; 25 j++; 26 } 27 } 28 29 return res; 30 } 31 };
sort nums1, 然后nums2 array 每一个element在 sorted 上做binary search.
Time Complexity: O(mlogm + nlogm), m = nums1.length, n = nums2.length.
Space: O(resArr.length).
AC Java:
1 public class Solution { 2 public int[] intersection(int[] nums1, int[] nums2) { 3 HashSet<Integer> res = new HashSet<Integer>(); 4 Arrays.sort(nums1); 5 for(int num : nums2){ 6 if(binarySearch(nums1, num)){ 7 res.add(num); 8 } 9 } 10 11 int [] resArr = new int[res.size()]; 12 int i = 0; 13 for(int num : res){ 14 resArr[i++] = num; 15 } 16 return resArr; 17 } 18 19 private boolean binarySearch(int [] nums, int target){ 20 int low = 0; 21 int high = nums.length-1; 22 while(low <= high){ 23 int mid = low + (high-low)/2; 24 if(nums[mid] < target){ 25 low = mid+1; 26 }else if(nums[mid] > target){ 27 high = mid-1; 28 }else{ 29 return true; 30 } 31 } 32 return false; 33 } 34 }
跟上Intersection of Two Arrays II, Intersection of Three Sorted Arrays.
