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[LeetCode] 349. Intersection of Two Arrays 两个数组相交
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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.
解法：由于结果中要求元素是唯一的，所以用set来统计num1 中的数字。再循环num2中的数字，在set中存在就记录到结果中，同时从set中删除。
Java: HashSet, T: O(n)
class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
HashSet<Integer> set = new HashSet<Integer>();
ArrayList<Integer> res = new ArrayList<Integer>();
//Add all elements to set from array 1
for(int i =0; i< nums1.length; i++) set.add(nums1[i]);
for(int j = 0; j < nums2.length; j++) {
// If present in array 2 then add to res and remove from set
if(set.contains(nums2[j])) {
res.add(nums2[j]);
set.remove(nums2[j]);
}
}
// Convert ArrayList to array
int[] arr = new int[res.size()];
for (int i= 0; i < res.size(); i++) arr[i] = res.get(i);
return arr;
}
}
Java: Hashset T: O(n)
public class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
Set<Integer> set = new HashSet<>();
Set<Integer> intersect = new HashSet<>();
for (int i = 0; i < nums1.length; i++) {
set.add(nums1[i]);
}
for (int i = 0; i < nums2.length; i++) {
if (set.contains(nums2[i])) {
intersect.add(nums2[i]);
}
}
int[] result = new int[intersect.size()];
int i = 0;
for (Integer num : intersect) {
result[i++] = num;
}
return result;
}
}
Java: two points, T: O(nlogn)
public class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
Set<Integer> set = new HashSet<>();
Arrays.sort(nums1);
Arrays.sort(nums2);
int i = 0;
int j = 0;
while (i < nums1.length && j < nums2.length) {
if (nums1[i] < nums2[j]) {
i++;
} else if (nums1[i] > nums2[j]) {
j++;
} else {
set.add(nums1[i]);
i++;
j++;
}
}
int[] result = new int[set.size()];
int k = 0;
for (Integer num : set) {
result[k++] = num;
}
return result;
}
}
Java: Binary Search, T: O(nlogn)
public class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
Set<Integer> set = new HashSet<>();
Arrays.sort(nums2);
for (Integer num : nums1) {
if (binarySearch(nums2, num)) {
set.add(num);
}
}
int i = 0;
int[] result = new int[set.size()];
for (Integer num : set) {
result[i++] = num;
}
return result;
}
public boolean binarySearch(int[] nums, int target) {
int low = 0;
int high = nums.length - 1;
while (low <= high) {
int mid = low + (high - low) / 2;
if (nums[mid] == target) {
return true;
}
if (nums[mid] > target) {
high = mid - 1;
} else {
low = mid + 1;
}
}
return false;
}
}
Python:
class Solution(object):
def intersection(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
res = []
s = set()
for num in nums1:
s.add(num) for num in nums2:
if num in s:
res.append(num)
s.remove(num) return res
Python:
class Solution(object):
def intersection(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
nums1=set(nums1)
nums2=set(nums2)
return list(nums1&nums2)
C++:
class Solution {
public:
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
set<int> s(nums1.begin(), nums1.end()), res;
for (auto a : nums2) {
if (s.count(a)) res.insert(a);
}
return vector<int>(res.begin(), res.end());
}
};
C++:
class Solution {
public:
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
vector<int> res;
int i = 0, j = 0;
sort(nums1.begin(), nums1.end());
sort(nums2.begin(), nums2.end());
while (i < nums1.size() && j < nums2.size()) {
if (nums1[i] < nums2[j]) ++i;
else if (nums1[i] > nums2[j]) ++j;
else {
if (res.empty() || res.back() != nums1[i]) {
res.push_back(nums1[i]);
}
++i; ++j;
}
}
return res;
}
};
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[LeetCode] 350. Intersection of Two Arrays II 两个数组相交II
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