https://leetcode.com/problems/intersection-of-two-arrays-ii/
https://leetcode.com/problems/intersection-of-two-arrays-ii/discuss/82238/4ms-java-solution
https://www.programcreek.com/2014/05/leetcode-intersection-of-two-arrays-ii-java/
https://github.com/mintycc/OnlineJudge-Solutions/blob/master/Leetcode/349_Intersection_of_Two_Arrays.java
Given two arrays, write a function to compute their intersection.
Example 1:
Input: nums1 = [1,2,2,1], nums2 = [2,2] Output: [2,2]
Example 2:
Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4] Output: [4,9]
Note:
- Each element in the result should appear as many times as it shows in both arrays.
- The result can be in any order.
Follow up:
- What if the given array is already sorted? How would you optimize your algorithm?
- What if nums1's size is small compared to nums2's size? Which algorithm is better?
- What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?
public int[] intersect(int[] nums1, int[] nums2) {
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
ArrayList<Integer> result = new ArrayList<Integer>();
for(int i = 0; i < nums1.length; i++)
{
if(map.containsKey(nums1[i])) map.put(nums1[i], map.get(nums1[i])+1);
else map.put(nums1[i], 1);
}
for(int i = 0; i < nums2.length; i++)
{
if(map.containsKey(nums2[i]) && map.get(nums2[i]) > 0)
{
result.add(nums2[i]);
map.put(nums2[i], map.get(nums2[i])-1);
}
}
int[] r = new int[result.size()];
for(int i = 0; i < result.size(); i++)
{
r[i] = result.get(i);
}
return r;
}https://leetcode.com/problems/intersection-of-two-arrays-ii/discuss/82238/4ms-java-solution
Arrays.sort(nums1);
Arrays.sort(nums2);
int pnt1 = 0;
int pnt2 = 0;
ArrayList<Integer> myList = new ArrayList<Integer>();
while((pnt1 < nums1.length) &&(pnt2< nums2.length)){
if(nums1[pnt1]<nums2[pnt2]){
pnt1++;
}
else{
if(nums1[pnt1]>nums2[pnt2]){
pnt2++;
}
else{
myList.add(nums1[pnt1]);
pnt1++;
pnt2++;
}
}
}
int[] res = new int[myList.size()];
for(int i = 0; i<res.length; i++){
res[i] = (Integer)myList.get(i);
}
return res;https://www.programcreek.com/2014/05/leetcode-intersection-of-two-arrays-ii-java/
https://github.com/mintycc/OnlineJudge-Solutions/blob/master/Leetcode/349_Intersection_of_Two_Arrays.java
1. `List<Integer> ans = new ArrayList<Integer>();`
2. `l = mid + 1;` -> `int mid = (l + r) / 2 - 1;`
3. return value
4. binarySearch nums2[] maybe null
*/
class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
Arrays.sort(nums1);
Arrays.sort(nums2);
List<Integer> ans = new ArrayList<Integer>();
for (int i = 0; i < nums1.length; i ++) {
if (binarySearch(nums2, nums1[i]))
ans.add(nums1[i]);
while (i + 1 < nums1.length && nums1[i + 1] == nums1[i]) i ++;
}
int[] rtn = new int[ans.size()];
for (int i = 0; i < ans.size(); i ++)
rtn[i] = ans.get(i);
return rtn;
}
private boolean binarySearch(int[] nums, int target) {
if (nums.length == 0) return false;
int l = 0, r = nums.length;
while (l < r - 1) {
int mid = (l + r) / 2 - 1;
if (nums[mid] == target) return true;
if (nums[mid] < target) l = mid + 1;
else r = mid;
}
return nums[l] == target;
}
}
class Solution {
public int[] intersection(int[] nums1, int[] nums2) {
Set<Integer> set = new HashSet<>();
for (int num : nums1)
set.add(num);
Set<Integer> ans = new HashSet<>();
for (int num : nums2)
if (set.contains(num))
ans.add(num);
int[] rtn = new int[ans.size()];
int count = 0;
for (int num : ans)
rtn[count ++] = num;
return rtn;
}
public int[] intersect(int[] nums1, int[] nums2) {
Arrays.sort(nums1);
Arrays.sort(nums2);
int i = 0, j = 0;
List<Integer> ans = new ArrayList<>();
while (i < nums1.length && j < nums2.length)
if (nums1[i] < nums2[j]) i ++;
else if (nums1[i] > nums2[j]) j ++;
else {
ans.add(nums1[i]);
i ++; j ++;
}
int[] rtn = new int[ans.size()];
for (i = 0; i < ans.size(); i ++)
rtn[i] = ans.get(i);
return rtn;
}
}
// border + BinarySearch
class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
Arrays.sort(nums1);
Arrays.sort(nums2);
int floor = 0;
List<Integer> ans = new ArrayList<>();
for (int num : nums1) {
int tmp = binarySearch(nums2, num, floor);
if (tmp > -1) {
ans.add(num);
floor = tmp + 1;
}
}
int[] rtn = new int[ans.size()];
for (int i = 0; i < ans.size(); i ++)
rtn[i] = ans.get(i);
return rtn;
}
private int binarySearch(int[] nums, int target, int floor) {
if (floor >= nums.length) return -1;
int l = floor, r = nums.length;
while (l < r - 1) {
int mid = l + (r - l) / 2 - 1;
if (nums[mid] < target) l = mid + 1;
else r = mid + 1;
}
if (nums[l] == target) return l;
return -1;
}
1. 如果两个array分别是[1, 2, 3, 4, .... 10000]以及[1, 2, 4, 10001]， 如果
⽤用两个指针遍历runtime是10000， 事实上intersection只出现在前四个位
置，怎么improve
2. 有重复元素和⽆无重复元素版本
3. ⼀一个数组⼤大，⼀一个数组⼩小
4. 两个数组都很⼤大
5. 这题和leetcode原题还不不⼀一样。这题要求输出所有inersection, 不不能去
重。⽤用two pointer 做可以搞定。最后她加我⽤用⼆二分法写⼀一遍，然后我卡
在加⼊入重复这⾥里里。我给的思路路是找lower bound，然后两个数组同时向前
移动。貌似也不不是她想要得，她最后给出的解答是每⼀一次找到相同的元
素，更更新⼆二分区间到找到的元素下⼀一个到数组结尾。
6. https://articles.leetcode.com/here-is-phone-screening-question-from/
