349 sum of intersection of two arrays and 01

One 、01 Sum of two numbers

1.1 subject

Given an array of integers nums And a target value target, Please find and as the target value in the array Two Integers , And return their array subscripts .
You can assume that each input corresponds to only one answer . however , The same element in an array cannot be used twice .
Example :

Given nums = [2, 7, 11, 15], target = 9
because nums[0] + nums[1] = 2 + 7 = 9
So back [0, 1]

analysis ：
The meaning of the question is to let you find two positions in the array, and the sum of their corresponding positions is target.

1.2 Realization

public class TwoSum {
    
    public static void main(String[] args) {
    
        int[] nums1 = {
    4,9,5};
        int[] res = twoSum(nums1, 9);
        System.out.println(res[0]);
        System.out.println(res[1]);
    }

    public static int[] twoSum(int[] nums, int target) {
    
        // utilize map structure ,key Store values ,value Store subscripts 
        int[] res = new int[2];
        if(nums == null || nums.length == 0){
    
            return res;
        }
        Map<Integer, Integer> map = new HashMap<>();
        for(int i = 0; i < nums.length; i++){
    
            int temp = target - nums[i];
            if(map.containsKey(temp)){
    
                res[1] = i;
                res[0] = map.get(temp);
            }
            map.put(nums[i], i);
        }
        return res;
    }

}

Two 、01 Sum of two numbers

2.1 subject

Given two arrays , Write a function to calculate their intersection .
Example 1:
Input : nums1 = [1,2,2,1], nums2 = [2,2]
Output : [2]
Example 1:
Input : nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output : [9,4]
explain :
Each element of the output must be unique .
We can ignore the order of output results .

2.2 Realization

/**349.  Intersection of two arrays **/
public class Intersection {
    
    /** *  establish set aggregate , Use its weight removal characteristics to screen  *  establish nums aggregate , Used to store elements that do not overlap  *  Loop through groups 2, If set Contains an array 2 The elements of , Then add this element to set2 in  *  Traverse set aggregate , Store the results in an array  */
    public static void main(String[] args) {
    
        int[] nums1 = {
    4,9,5};
        int[] nums2 = {
    9,4,9,8,4};
        System.out.println(Arrays.toString(intersection2(nums1, nums2)));
    }

    //  Method 1：Set aggregate 
    public static int[] intersection1(int[] nums1, int[] nums2) {
    
        Set<Integer> set1 = new HashSet<>();    // HashSet characteristic ： disorder 、 No repetition 、 No index .
        Set<Integer> set2 = new HashSet<>();

        for(int i = 0; i < nums1.length; i++){
    
            set1.add(nums1[i]);
        }

        for(int i = 0; i < nums2.length; i++){
    
            if(set1.contains(nums2[i])){
    
                set2.add(nums2[i]);
            }
        }

        int[] nums = new int[set2.size()];
        int k = 0;
        for(Integer it : set2){
    
            nums[k++] = it;
        }
        return nums;
    }

    //  Method 2：Map aggregate 
    public static int[] intersection2(int[] nums1, int[] nums2) {
    
        //  establish map aggregate 
        Map<Integer, Integer> map = new HashMap<>();
        //  Will array 1 All the elements in are added to map Collection , And the default value is 1, It is equivalent to the operation of weight removal 
        for(int i = 0; i < nums1.length; i++){
    
            map.put(nums1[i], 1);
        }
        //  establish list aggregate , Used to store results 
        ArrayList<Integer> list = new ArrayList<>();
        //  Loop through groups 2, If map Contains arrays 2 The elements of , And its value Greater than 0, Then add it to list Collection , Update it value
        for(int j = 0; j < nums2.length; j++){
    
            if(map.containsKey(nums2[j]) && map.get(nums2[j]) > 0){
    
                list.add(nums2[j]);
                map.put(nums2[j], map.get(nums2[j]) - 1);//  Put it value to update , That is to say -1, It's equivalent to removing weight 
            }
        }
        //  Create array , For storing results , take list The elements in are assigned to the array , And back to 
        int len = list.size();
        int[] res = new int[len];
        for(int i = 0; i < len; i++){
    
            res[i] = list.remove(0);
        }
        return res;
    }
}