Question

350. Intersection of Two Arrays II

中文文档

Description

Given two integer arrays nums1 and nums2, return _an array of their intersection_. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.

 

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]
Explanation: [9,4] is also accepted.

 

Constraints:

• 1 <= nums1.length, nums2.length <= 1000
• 0 <= nums1[i], nums2[i] <= 1000

 

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?

Solutions

Solution 1

class Solution:
  def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
    counter = Counter(nums1)
    res = []
    for num in nums2:
      if counter[num] > 0:
        res.append(num)
        counter[num] -= 1
    return res

class Solution {
  public int[] intersect(int[] nums1, int[] nums2) {
    Map<Integer, Integer> counter = new HashMap<>();
    for (int num : nums1) {
      counter.put(num, counter.getOrDefault(num, 0) + 1);
    }
    List<Integer> t = new ArrayList<>();
    for (int num : nums2) {
      if (counter.getOrDefault(num, 0) > 0) {
        t.add(num);
        counter.put(num, counter.get(num) - 1);
      }
    }
    int[] res = new int[t.size()];
    for (int i = 0; i < res.length; ++i) {
      res[i] = t.get(i);
    }
    return res;
  }
}

class Solution {
public:
  vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
    unordered_map<int, int> counter;
    for (int num : nums1) ++counter[num];
    vector<int> res;
    for (int num : nums2) {
      if (counter[num] > 0) {
        --counter[num];
        res.push_back(num);
      }
    }
    return res;
  }
};

func intersect(nums1 []int, nums2 []int) []int {
  counter := make(map[int]int)
  for _, num := range nums1 {
    counter[num]++
  }
  var res []int
  for _, num := range nums2 {
    if counter[num] > 0 {
      counter[num]--
      res = append(res, num)
    }
  }
  return res
}

function intersect(nums1: number[], nums2: number[]): number[] {
  const map = new Map<number, number>();
  for (const num of nums1) {
    map.set(num, (map.get(num) ?? 0) + 1);
  }

  const res = [];
  for (const num of nums2) {
    if (map.has(num) && map.get(num) !== 0) {
      res.push(num);
      map.set(num, map.get(num) - 1);
    }
  }
  return res;
}

use std::collections::HashMap;
impl Solution {
  pub fn intersect(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
    let mut map = HashMap::new();
    for num in nums1.iter() {
      *map.entry(num).or_insert(0) += 1;
    }

    let mut res = vec![];
    for num in nums2.iter() {
      if map.contains_key(num) && map.get(num).unwrap() != &0 {
        map.insert(num, map.get(&num).unwrap() - 1);
        res.push(*num);
      }
    }
    res
  }
}

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersect = function (nums1, nums2) {
  const counter = {};
  for (const num of nums1) {
    counter[num] = (counter[num] || 0) + 1;
  }
  let res = [];
  for (const num of nums2) {
    if (counter[num] > 0) {
      res.push(num);
      counter[num] -= 1;
    }
  }
  return res;
};

public class Solution {
  public int[] Intersect(int[] nums1, int[] nums2) {
    HashSet<int> hs1 = new HashSet<int>(nums1.Concat(nums2).ToArray());
    Dictionary<int, int> dict = new Dictionary<int, int>();
    List<int> result = new List<int>();

    foreach (int x in hs1) {
      dict[x] = 0;
    }

    foreach (int x in nums1) {
      if (dict.ContainsKey(x)) {
        dict[x] += 1;
      } else {
        dict[x] = 1;
      }
    }

    foreach (int x in nums2) {
      if (dict[x] > 0) {
        result.Add(x);
        dict[x] -=1;
      }
    }

    return result.ToArray();
  }
}

class Solution {
  /**
   * @param Integer[] $nums1
   * @param Integer[] $nums2
   * @return Integer[]
   */
  function intersect($nums1, $nums2) {
    $rs = [];
    for ($i = 0; $i < count($nums1); $i++) {
      $hashtable[$nums1[$i]] += 1;
    }
    for ($j = 0; $j < count($nums2); $j++) {
      if (isset($hashtable[$nums2[$j]]) && $hashtable[$nums2[$j]] > 0) {
        array_push($rs, $nums2[$j]);
        $hashtable[$nums2[$j]] -= 1;
      }
    }
    return $rs;
  }
}