1360. Dates are a few days apart

Please write a program to calculate how many days are between two dates.

The date is given as a string in the format YYYY-MM-DD, as shown in the example.

Example 1:

Input: date1 = "2019-06-29", date2 = "2019-06-30"
Output: 1

This problem is lazy and solved directly by using Java library functions.(

import java.time.LocalDate;
import java.time.temporal.ChronoUnit;

class Solution {
    public int daysBetweenDates(String date1, String date2) {
        LocalDate startDate = LocalDate.parse(date1);
        LocalDate endDate = LocalDate.parse(date2);
        long daysDiff = ChronoUnit.DAYS.between(startDate, endDate);
        return Math.abs((int)daysDiff);
    }
}

1361. Verify Binary Tree

There are n nodes i n the binary tree, numbered from 0 to n-1, where the two children of node i are leftChild[i] and rightChild[i].

Returns true only if all nodes can form and only form a valid binary tree; otherwise returns false.

If node i has no left child node, then leftChild[i] equals -1.The right child node also conforms to this rule.

Note: Node has no value, only node number is used in this problem.

Example 1:

Input: n = 4, leftChild = [1,-1,3,-1], rightChild = [2,-1,-1,-1]
Output:true

Be careful:

Sufficient conditions for binary trees:

1) Root node is not accessible

2) Other nodes are accessed only once

class Solution {
    public boolean validateBinaryTreeNodes(int n, int[] leftChild, int[] rightChild) {
        int[] cnt = new int[n];
        for(int i=0; i<n; i++) {
            if(leftChild[i] >= 0) {
                cnt[leftChild[i]] += 1;
            }
            if(rightChild[i] >= 0) {
                cnt[rightChild[i]] += 1;
            }
        }
        if(cnt[0] != 0) {
            return false;
        }
        for(int i=1; i<n; i++) {
            if(cnt[i] != 1) {
                return false;
            }
        }
        return true;
    }
}

1362. Closest factor

Give you an integer, num, and find two integers that meet all of the following requirements:

The product of two numbers equals num + 1 or num + 2
Measured by absolute difference, the two numbers are closest
You can return these two integers in any order.

Example 1:

Input: num = 8
 Output: [3,3]
Interpretation: For num + 1 = 9, the closest two factors are 3 & 3; for num + 2 = 10, the closest two factors are 2 & 5, thus returning 3 & 3.

Skill: Traversing from sqrt to 1, the difference must be increasing gradually, so the first difference you encounter must be the smallest.

class Solution {
    public int[] closestDivisors(int num) {
        int[] r = new int[2];
        int sqrt = (int) Math.sqrt(num + 2);  //square root
        for(int i = sqrt; i >= 1; i--){
            if((num + 1) % i == 0){
                r[0] = i;
                r[1] = (num + 1) / i;
                return r;
            }
            if((num + 2) % i == 0){
                r[0] = i;
                r[1] = (num + 2) / i;
                return r;
            }
        }
        return r;
    }
}

1363. Form a maximum multiple of three

Give you an array of integers, digits. You can form a multiple of 3 by joining some of the numbers in any order. Please return the maximum multiple of 3 you can get.

Since the answer may not be in the range of integer data types, return the answer as a string.

If you cannot get an answer, return an empty string.

Example 1:

Input: digits = [8,1,9]
Output:'981'

Analysis:

Divided by 3, the sum of digits in each bit can also be divided by 3.

• Calculates the sum of digit arrays.
• Divide the numbers into 3i, 3i+1, 3i+2.
• If sum% 3 == 1, delete a 3i+1. If there is no such number, delete two 3i+2 (there are at least two 3i+2, as you can see); if sum% 3 == 2, the same is true.

The size of the observed data is only 0 to 9, and the number of occurrences of each number can be counted.

class Solution {
    private int[] num;

    private boolean delete(int k) {
        for (int i = k; i < 10; i = i + 3) {
            if (num[i] > 0) {
                num[i]--;
                return true;
            }
        }
        return false;
    }
    private String makeString() {
        StringBuilder ans = new StringBuilder("");
        for (int i = 9; i >= 0; i--)
            while (num[i]-- > 0)
                ans.append((char)(i + '0'));
        int index = 0;
        while (index < ans.length() - 1) {
            if (ans.charAt(index) != '0') break;
            index++;
        }
        return ans.delete(0, index).toString();
    }
    public String largestMultipleOfThree(int[] digits) {
        num = new int[10];
        int sum = 0;
        for (int value : digits) {
            num[value]++;
            sum += value;
        }
        int k = sum % 3;
        if (k > 0) {
            if (!delete(k)) {
                delete(3 - k);
                delete(3 - k);
            }
        }
        return makeString();
    }
}