LeetCode.349(350) Intersection of Two Arrays && II

题目349：

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

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

Note:

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

分析349(简单-易理解):

class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        //给定两个数组，求出两者交互的部分，其中返回其中结果不包含重复值，对顺序不做要求
        //思路：使用两个set来求，一个参考集，一个作为结果集
        HashSet<Integer> set1=new HashSet<Integer>();
        HashSet<Integer> set2=new HashSet<Integer>();
        
        //参考集合
        for(int i=0;i<nums1.length;i++){
            set1.add(nums1[i]);
        }
        
        //查找结果集
        for(int i=0;i<nums2.length;i++){
            if(set1.contains(nums2[i])){
                set2.add(nums2[i]);
            }
        }
        //将结果集合转成数组
        int [] res=new int[set2.size()];
        int count=0;
        for(int n:set2){
            res[count++]=n;
        }
        
        return res;
    }
}

分析349（参考答案）：

class Solution {
       public int[] intersection(int[] nums1, int[] nums2) {
         //求两个数组中相同的数，不包含重复值
        //参考答案
         //思路：利用两个数组实现Set的作用，分别标记1和2来实现不添加重复元素
           
        int max = 0;
        for (int i = 0; i < nums1.length; i++) {
            max = Math.max(max, nums1[i]);
        }
        
        for (int i = 0; i < nums2.length; i++) {
            max = Math.max(max, nums2[i]);
        }
        
        int[] arr = new int[max + 1];
        for (int i = 0; i < nums1.length; i++) {
            int n = nums1[i];
            arr[n] = 1;
        }
        int count = 0;
        for (int i = 0; i < nums2.length; i++) {
            int n = nums2[i];
            if(arr[n] == 1) {
                //更改2的作用是再次不在添加相同的数据
                arr[n] = 2;
                count++;
            }
        }
        
        int[] result = new int[count];
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == 2) {
                //分别给数据赋值
                result[--count] = i;
            }
        }
        return result;
    }
}

题目350:

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

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

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?

分析：

class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        //类似347,只是这次允许重复数据
        //要求：数据有序，nums1的长度小nums2的长度，nums2的数据在硬盘中，不内存不允许将所有数据取出
        //思路：使用HashMap来存储nums1的所有元素，通过计数来判断nums1的参考元素是否用完，用完则结束匹配
        
        int count=nums1.length;
        HashMap<Integer,Integer> hm=new HashMap<Integer,Integer>();
        List<Integer> list=new ArrayList<Integer>();
        for(int i=0;i<nums1.length;i++){
            //将元素存入
            hm.put(nums1[i],hm.getOrDefault(nums1[i],0)+1);
        }
        
        for(int i=0;i<nums2.length&&count>0;i++){
            if(hm.containsKey(nums2[i])&&hm.get(nums2[i])>0){
                hm.put(nums2[i],hm.get(nums2[i])-1);
                //记录该元素
                list.add(nums2[i]);
                count--;
            }
        }
        
        //返回结果
        int i=0;
        int [] res=new int[list.size()];
        for(int n:list){
            res[i++]=n;
        }
        return res;
    }
}