[LeetCode] 349. Intersection of Two Arrays 两个数组相交

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

[LeetCode] 160. Intersection of Two Linked Lists 求两个链表的交集

　　

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