leetcode 350: 两个数组的交集II

package com.example.lettcode.dailyexercises;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

/**
 * @Class Intersect
 * @Description 350 两个数组的交集II
 * 给定两个数组，编写一个函数来计算它们的交集。
 * <p>
 * 示例 1:
 * 输入: nums1 = [1,2,2,1], nums2 = [2,2]
 * 输出: [2,2]
 * <p>
 * 示例 2:
 * 输入: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
 * 输出: [4,9]
 * <p>
 * 说明：
 * 我们可以不考虑输出结果的顺序。
 * @Author
 * @Date 2020/7/13
 **/
public class Intersect {   
}

/**
 * 解法1:利用hash
 */
public static int[] intersect(int[] nums1, int[] nums2) {
	if (nums1 == null || nums1.length == 0
			|| nums2 == null || nums2.length == 0) return new int[]{};
	List<Integer> ans = new ArrayList<>();
	Map<Integer, Integer> integerMap = new HashMap<>();
	for (int num : nums1) {
		if (integerMap.containsKey(num)) {
			int count = integerMap.get(num);
			count++;
			integerMap.put(num, count);
		} else {
			integerMap.put(num, 1);
		}
	}

	for (int num : nums2) {
		if (integerMap.containsKey(num)) {
			ans.add(num);
			int count = integerMap.get(num);
			count--;
			if (count == 0) {
				integerMap.remove(num);
				continue;
			}
			integerMap.put(num, count);
		}
	}
	return ans.stream().mapToInt(Integer::valueOf).toArray();
}

/**
 * 解法2:利用排序,时间复杂度为O(NlgN)
 */
public static int[] intersect(int[] nums1, int[] nums2) {
	if (nums1 == null || nums1.length == 0
			|| nums2 == null || nums2.length == 0) return new int[]{};
	quickSort(nums1, 0, nums1.length - 1);
	quickSort(nums2, 0, nums2.length - 1);
	int p = 0, q = 0;
	List<Integer> integerList = new ArrayList<>();
	while (p < nums1.length && q < nums2.length) {
		if (nums1[p] > nums2[q]) q++;
		else if (nums1[p] < nums2[q]) p++;
		else {
			integerList.add(nums1[p]);
			p++;
			q++;
		}
	}
	return integerList.stream().mapToInt(Integer::valueOf).toArray();
}

public static void quickSort(int[] num, int start, int end) {
	if (num == null || num.length <= 1) return;
	if (start > end) return;
	int temp = num[start];
	int p = start, q = end;
	while (p < q) {
		while (p < q && temp <= num[q]) q--;
		while (p < q && temp >= num[p]) p++;

		if (p < q) {
			num[p] += num[q];
			num[q] = num[p] - num[q];
			num[p] = num[p] - num[q];
		}
	}
	num[start] = num[p];
	num[p] = temp;
	quickSort(num, start, p - 1);
	quickSort(num, p + 1, end);
}

// 测试用例
public static void main(String[] args) {
	int[] num1 = new int[]{1, 2, 2, 1};
	int[] num2 = new int[]{2, 2};
	int[] ans = intersect(num1, num2);
	System.out.println("Intersect result demo01:");
	for (int num : ans) {
		System.out.print("," + num);
	}
	System.out.println("");

	num1 = new int[]{4, 9, 5};
	num2 = new int[]{9, 4, 9, 8, 4};
	ans = intersect(num1, num2);
	System.out.println("Intersect result demo02:");
	for (int num : ans) {
		System.out.print("," + num);
	}
	System.out.println("");
}

leetcode 349: 两个数组的交集
https://www.cnblogs.com/fyusac/p/13913356.html