LeetCode - Easy - 350. Intersection of Two Arrays II

灰太狼 2023-01-10 11:50 183阅读 0赞

## Topic ##

*  Hash Table
 *  Two Pointers
 *  Binary Search
 *  Sort

## Description ##

[https://leetcode.com/problems/intersection-of-two-arrays-ii/][https_leetcode.com_problems_intersection-of-two-arrays-ii]

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?

## Analysis ##

方法一：我写的。

方法二：别人写的，与方法一思路一致，代码简洁些。

方法三：排序两数组，然后用两指针齐齐比较得出结果。

方法四：Java8式写法。

## Submission ##

import java.util.ArrayList;
    import java.util.Arrays;
    import java.util.HashMap;
    import java.util.List;
    import java.util.Map;
    import java.util.stream.Collectors;
    
    public class IntersectionOfTwoArraysII { 
    
    	// 方法一：我写的
    	public int[] intersect1(int[] nums1, int[] nums2) { 
    		if (nums1 == null || nums2 == null || //
    				nums1.length == 0 || nums2.length == 0)
    			return new int[] { };
    
    		if (nums1.length > nums2.length) { // 这if语块可省略
    			int[] temp = nums1;
    			nums1 = nums2;
    			nums2 = temp;
    		}
    
    		Map<Integer, Integer> num2count = new HashMap<>();
    
    		for (int n : nums1) { 
    			Integer count = num2count.get(n);
    			if (count == null)
    				count = 0;
    			num2count.put(n, count + 1);
    		}
    
    		List<Integer> list = new ArrayList<>();
    		for (int n : nums2) { 
    			Integer count = num2count.get(n);
    			if (count != null && count > 0) { 
    				list.add(n);
    				num2count.put(n, count - 1);
    			}
    		}
    		int[] result = new int[list.size()];
    
    		for (int i = 0; i < list.size(); i++)
    			result[i] = list.get(i);
    
    		return result;
    	}
    
    	// 方法二：别人写的，与方法一思路一致，代码简洁些
    	public int[] intersect2(int[] nums1, int[] nums2) { 
    		HashMap<Integer, Integer> map = new HashMap<>();
    		for (int i : nums1)
    			map.put(i, map.getOrDefault(i, 0) + 1);
    
    		ArrayList<Integer> list = new ArrayList<>();
    		for (int i : nums2) { 
    			if (map.get(i) != null && map.get(i) > 0) { 
    				list.add(i);
    				map.put(i, map.get(i) - 1);
    			}
    		}
    		int[] ret = new int[list.size()];
    		for (int i = 0; i < list.size(); i++) { 
    			ret[i] = list.get(i);
    		}
    		return ret;
    	}
    
    	// 方法三：排序两数组，然后齐齐
    	public int[] intersect3(int[] nums1, int[] nums2) { 
    		Arrays.sort(nums1);
    		Arrays.sort(nums2);
    		int n = nums1.length, m = nums2.length;
    		int i = 0, j = 0;
    		List<Integer> list = new ArrayList<>();
    		while (i < n && j < m) { 
    			int a = nums1[i], b = nums2[j];
    			if (a == b) { 
    				list.add(a);
    				i++;
    				j++;
    			} else if (a < b) { 
    				i++;
    			} else { 
    				j++;
    			}
    		}
    		int[] ret = new int[list.size()];
    		for (int k = 0; k < list.size(); k++)
    			ret[k] = list.get(k);
    		return ret;
    	}
    
    	// 方法四：Java8版
    	public int[] intersect4(int[] nums1, int[] nums2) { 
    		Map<Integer, Long> map = Arrays.stream(nums2).boxed()
    				.collect(Collectors.groupingBy(e -> e, Collectors.counting()));
    		return Arrays.stream(nums1).filter(e -> { 
    			if (!map.containsKey(e))
    				return false;
    			map.put(e, map.get(e) - 1);
    			if (map.get(e) == 0)
    				map.remove(e);
    			return true;
    		}).toArray();
    	}
    
    }

## Test ##

import static org.junit.Assert.*;
    
    import java.util.Arrays;
    
    import org.junit.Test;
    
    public class IntersectionOfTwoArraysIITest { 
    
    	@Test
    	public void test1() { 
    		IntersectionOfTwoArraysII obj = new IntersectionOfTwoArraysII();
    
    		int[] array1 = {  1, 2, 2, 1 };
    		int[] array2 = {  2, 2 };
    
    		assertArrayEquals(array2, obj.intersect1(array1, array2));
    		assertArrayEquals(array2, obj.intersect2(array1, array2));
    		assertArrayEquals(array2, obj.intersect3(array1, array2));
    		assertArrayEquals(array2, obj.intersect4(array1, array2));
    	}
    
    	@Test
    	public void test2() { 
    		IntersectionOfTwoArraysII obj = new IntersectionOfTwoArraysII();
    
    		int[] array1 = {  4, 9, 5 };
    		int[] array2 = {  9, 4, 9, 8, 4 };
    		int[] expected = {  9, 4 }, actual;
    		Arrays.sort(expected);
    
    		actual = obj.intersect1(array1, array2);
    		Arrays.sort(actual);
    		assertArrayEquals(expected, actual);
    
    		actual = obj.intersect2(array1, array2);
    		Arrays.sort(actual);
    		assertArrayEquals(expected, actual);
    
    		actual = obj.intersect3(array1, array2);
    		Arrays.sort(actual);
    		assertArrayEquals(expected, actual);
    
    		actual = obj.intersect4(array1, array2);
    		Arrays.sort(actual);
    		assertArrayEquals(expected, actual);
    
    	}
    }

[https_leetcode.com_problems_intersection-of-two-arrays-ii]: https://leetcode.com/problems/intersection-of-two-arrays-ii/