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.

 

M1: binary search

先把nums2排序，然后对于nums1中的每一个元素，在nums2中进行binary search，需要一个set存结果

time: O(nlogn), space: O(1)

class Solution {    public int[] intersection(int[] nums1, int[] nums2) {        Set
     
       set = new HashSet<>();        Arrays.sort(nums2);        for(int n : nums1) {            if(binarySearch(nums2, n)) {                set.add(n);            }        }                int[] res = new int[set.size()];        int i = 0;        for(int n : set) {            res[i++] = n;        }        return res;    }        public boolean binarySearch(int[] nums, int target) {        int left = 0, right = nums.length - 1;        while(left <= right) {            int mid = left + (right - left) / 2;            if(nums[mid] == target) {                return true;            } else if(nums[mid] > target) {                right = mid - 1;            } else {                left = mid + 1;            }        }        return false;    }}
     

 

 

M2: hash table

用两个set，先把nums1元素放进set1，如果nums2中也有该元素，放进res set里

time: O(n), space: O(n)

class Solution {    public int[] intersection(int[] nums1, int[] nums2) {        Set
     
       set1 = new HashSet<>();        Set
      
        set2 = new HashSet<>();        for(int n : nums1) {            set1.add(n);        }        for(int n : nums2) {            if(set1.contains(n)) {                set2.add(n);            }        }                int[] res = new int[set2.size()];        int i = 0;        for(int n : set2) {            res[i++] = n;        }        return res;    }}
      
     

 

M3: two pointers

需要先对两个数组都排序，并用set存结果去重

time: O(nlogn), space: O(1)

class Solution {    public int[] intersection(int[] nums1, int[] nums2) {        Set
     
       set = new HashSet<>();        Arrays.sort(nums1);        Arrays.sort(nums2);        int i = 0, j = 0;        while(i < nums1.length && j < nums2.length) {            if(nums1[i] == nums2[j]) {                set.add(nums1[i]);                i++;                j++;            } else if(nums1[i] > nums2[j]) {                j++;            } else {                i++;            }        }                int[] res = new int[set.size()];        int k = 0;        for(int n : set) {            res[k++] = n;        }        return res;    }}