547. Intersection of Two Arrays【easy】

Given two arrays, write a function to compute their intersection.

Notice

• Each element in the result must be unique.
• The result can be in any order.

Example

Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].

Challenge

Can you implement it in three different algorithms?

解法一：

 class Solution {
 public:
     /**
      * @param nums1 an integer array
      * @param nums2 an integer array
      * @return an integer array
      */
     vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
         sort(nums1.begin(), nums1.end());
         sort(nums2.begin(), nums2.end());

         vector<int> intersect;
         vector<int>::iterator it1 = nums1.begin(), it2 = nums2.begin();
         while ((it1 != nums1.end()) && (it2 != nums2.end())) {
             if (*it1 < *it2) {
                 it1++;
             } else if (*it1 > *it2) {
                 it2++;
             } else {
                 intersect.push_back(*it1);
                 it1++; it2++;
             }
         }

         auto last = unique(intersect.begin(), intersect.end());
         intersect.erase(last, intersect.end());
         return intersect;
     }
 };

sort & merge

类属性算法unique的作用是从输入序列中“删除”所有相邻的重复元素。该算法删除相邻的重复元素，然后重新排列输入范围内的元素，并且返回一个迭代器（容器的长度没变，只是元素顺序改变了），表示无重复的值范围得结束。在STL中unique函数是一个去重函数， unique的功能是去除相邻的重复元素(只保留一个),其实它并不真正把重复的元素删除，是把重复的元素移到后面去了，然后依然保存到了原数组中，然后 返回去重后最后一个元素的地址，因为unique去除的是相邻的重复元素，所以一般用之前都会要排一下序。

常见的用法如下：

 /*
  * sort words alphabetically so we can find the duplicates
  */
 sort(words.begin(), words.end()); 

 /* eliminate duplicate words:
  * unique reorders words so that each word appears once in the
  * front portion of words and returns an iterator one past the unique range;
  * erase uses a vector operation to remove the nonunique elements
  */
  vector<string>::iterator end_unique =  unique(words.begin(), words.end()); 

  words.erase(end_unique, words.end());

参考@NineChapter的代码

解法二：

 public class Solution {
     /**
      * @param nums1 an integer array
      * @param nums2 an integer array
      * @return an integer array
      */
     public int[] intersection(int[] nums1, int[] nums2) {
         Arrays.sort(nums1);
         Arrays.sort(nums2);

         int i = 0, j = 0;
         int[] temp = new int[nums1.length];
         int index = 0;
         while (i < nums1.length && j < nums2.length) {
             if (nums1[i] == nums2[j]) {
                 if (index == 0 || temp[index - 1] != nums1[i]) {
                     temp[index++] = nums1[i];
                 }
                 i++;
                 j++;
             } else if (nums1[i] < nums2[j]) {
                 i++;
             } else {
                 j++;
             }
         }

         int[] result = new int[index];
         for (int k = 0; k < index; k++) {
             result[k] = temp[k];
         }

         return result;
     }
 }

sort & merge

参考@NineChapter的代码

解法三：

 public class Solution {
     /**
      * @param nums1 an integer array
      * @param nums2 an integer array
      * @return an integer array
      */
     public int[] intersection(int[] nums1, int[] nums2) {
         if (nums1 == null || nums2 == null) {
             return null;
         }

         HashSet<Integer> hash = new HashSet<>();
         for (int i = 0; i < nums1.length; i++) {
             hash.add(nums1[i]);
         }

         HashSet<Integer> resultHash = new HashSet<>();
         for (int i = 0; i < nums2.length; i++) {
             if (hash.contains(nums2[i]) && !resultHash.contains(nums2[i])) {
                 resultHash.add(nums2[i]);
             }
         }

         int size = resultHash.size();
         int[] result = new int[size];
         int index = 0;
         for (Integer num : resultHash) {
             result[index++] = num;
         }

         return result;
     }
 }

hash map

参考@NineChapter的代码

解法四：

 public class Solution {
     /**
      * @param nums1 an integer array
      * @param nums2 an integer array
      * @return an integer array
      */
     public int[] intersection(int[] nums1, int[] nums2) {
         if (nums1 == null || nums2 == null) {
             return null;
         }

         HashSet<Integer> set = new HashSet<>();

         Arrays.sort(nums1);
         for (int i = 0; i < nums2.length; i++) {
             if (set.contains(nums2[i])) {
                 continue;
             }
             if (binarySearch(nums1, nums2[i])) {
                 set.add(nums2[i]);
             }
         }

         int[] result = new int[set.size()];
         int index = 0;
         for (Integer num : set) {
             result[index++] = num;
         }

         return result;
     }

     private boolean binarySearch(int[] nums, int target) {
         if (nums == null || nums.length == 0) {
             return false;
         }

         int start = 0, end = nums.length - 1;
         while (start + 1 < end) {
             int mid = (end - start) / 2 + start;
             if (nums[mid] == target) {
                 return true;
             }
             if (nums[mid] < target) {
                 start = mid;
             } else {
                 end = mid;
             }
         }

         if (nums[start] == target) {
             return true;
         }
         if (nums[end] == target) {
             return true;
         }

         return false;
     }
 }

sort & binary search

参考@NineChapter的代码

解法五：

 public class Solution {
     public int[] intersection(int[] nums1, int[] nums2) {
         // Write your code here
         HashSet<Integer> set1 = new HashSet<>();
         ArrayList<Integer> result = new ArrayList<>();
         for(int n1 : nums1){
             set1.add(n1);
         }
         for(int n2 : nums2){
             if(set1.contains(n2)){
                 result.add(n2);
                 set1.remove(n2);
             }
         }
         int[] ret = new int[result.size()];
         for(int i = 0; i < result.size(); i++){
             ret[i] = result.get(i);
         }
         return ret;
     }
 };

参考@NineChapter的代码