basic data structures

 

content

1. Intersection of two numbers

2. The best time to buy and sell stocks

3. Merge two arrays

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1. Intersection of two numbers

350. Intersection of Two Arrays II - LeetCode (leetcode-cn.com) https://leetcode-cn.com/problems/intersection-of-two-arrays-ii/

Topics and examples:

 Problem-solving ideas ①: (using sets to solve problems)

①Store the number in nums1 into list1

② Determine whether the number in nums2 exists in list1, and if so, store the number in list2

③ At the same time, delete the number in list1

code show as below:

class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
List<Integer>list1=new ArrayList<>();
List<Integer>list2=new ArrayList<>();
for(int i=0;i<nums1.length;i++){
    list1.add(nums1[i]);
}  
 for (int num : nums2) {
            if (list1.contains(num)) {
                list2.add(num);
                // 从 list1 除去已匹配的数值
                list1.remove(Integer.valueOf(num));
            }
        }
int []tmp=new int [list2.size()];
int m=0;
for(int nums:list2){
    tmp[m++]=nums;
}return tmp;
}
    }

Problem-solving idea ②: (using hash map to solve the problem)

①Compare the lengths of nums1 and nums2, if nums1.length<nums2.length, then exchange the two arrays

②Using the correspondence between the two values ​​of Map<key,value>, store each element in nums1 into the hash Map, count with count=value, if the same key appears, count++;

③Use tmp=0; to record the subscript in the intersection

④ Traverse the array nums2, if there is an element contained in nums1, and the count>0 represented by it, copy the element to num1[tmp], and at the same time let tmp++ point to the next bit.

⑤ Let count--; reduce the number of times it appears in HashMap

⑥ Return the first k elements of nums1

The code is as follows: (The code segment has detailed comments for reference)

class Solution {
//交换数组的判断
    public int[] intersect(int[] nums1, int[] nums2) {
        if (nums1.length > nums2.length) {
            return intersect(nums2, nums1);
        }
//创建HashMap，并将nums1中的数存入其中
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int num : nums1) {
            int count = map.getOrDefault(num, 0) + 1;
            map.put(num, count);
        }
//创建一个新的等同nums1大小的数组
        int[] intersection = new int[nums1.length];
        int index = 0;
//判断在nums2中的数是否出现在HashMap中，并执行对应的方法
        for (int num : nums2) {
            int count = map.getOrDefault(num, 0);
            if (count > 0) {
                intersection[index++] = num;
                count--;
                if (count > 0) {
                    map.put(num, count);
                } else {
                    map.remove(num);
                }
            }
        }
//从下标from开始复制，复制到上标to，生成一个新的数组。注意这里包括下标from，不包括上标to。
        return Arrays.copyOfRange(intersection, 0, index);
    }
}

2. The best time to buy and sell stocks

121. Best time to buy and sell stocks - LeetCode (leetcode-cn.com) https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/

Topics and examples:

 Problem-solving ideas ①: (use the set method to solve)

Because the worst here may be 0, and the calculation is required to be the best case, the element non-repeatability of the set can be ignored here to complete

①Using the brute force solution, double loop, add each difference to the set

②When set is not equal to empty, directly use Collection.max(set) to find the maximum value in the set, because it is a reference value, so it needs to be unboxed and then returned

code show as below:

class Solution {
    public static int maxProfit(int[] prices) {
        Set<Integer> set = new HashSet<>();
        for (int i = 0; i < prices.length; i++) {
            for (int j = i + 1; j < prices.length; j++) {
                if (prices[i] < prices[j]) {
                    int m = prices[j] - prices[i];
                    set.add(m);
                }
            }
        }
        if (!set.isEmpty()) {
            Integer res = Collections.max(set);
            return res;
        } else {
            return 0;
        }
    }

Problem-solving ideas ②: (using dp dynamic programming to solve the problem)

①Find the minimum value before buying today

②Calculate the maximum profit of selling today

③Compare daily profits and get the maximum value

code show as below:

class Solution {
    public int maxProfit(int[] prices) {
        if(prices.length <= 1)
            return 0;
        int min = prices[0], max = 0;
        for(int i = 1; i < prices.length; i++) {
            max = Math.max(max, prices[i] - min);
            min = Math.min(min, prices[i]);
        }
        return max;
    }
}

3. Merge two arrays

88. Merge Two Sorted Arrays - LeetCode (leetcode-cn.com) https://leetcode-cn.com/problems/merge-sorted-array/ 

Topics and examples:

Problem-solving ideas: (In the previous sorting (2), there is an explanation in the merge sort)

class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        int s1=0;
        int e1=m-1;
        int s2=0;
        int e2=n-1;
        int index=0;
        int []tmp=new int[m+n];
        while (s1 <= e1 && s2 <= e2) {
            if(nums1[s1] <= nums2[s2]) {
                tmp[index] = nums1[s1];
                index++;
                s1++;
            }else {
                tmp[index] = nums2[s2];
                index++;
                s2++;
            }
        }
//此时表示s2数组已经走完，把剩下的s1数组中的数依次放入合并的数组中即可
        while (s1 <= e1) {
            tmp[index++] = nums1[s1++];
            //此处也可以像上面那样写成index++；s1++;
        }
//此时表示s1数组已经走完，把剩下的s2数组中的数依次放入合并的数组中即可
        while (s2 <= e2) {
            tmp[index++] = nums2[s2++];
        }
       for(int i=0;i<m+n;i++){
        nums1[i]=tmp[i];
       }
    }
    }

Come on duck~