Leetcode-2221: triangular sum of arrays

Title Description

I'll give you a subscript from 0 The starting array of integers nums , among nums[i] yes 0 To 9 Between （ Both contain ） A number of .
nums The values of the following elements are the final and subsequent operations ：
nums Initially contains n Elements . If n == 1 , Termination operation . otherwise , Create a new subscript from 0 The starting length is n - 1 Array of integers for newNums .
For satisfying 0 <= i < n - 1 The subscript i ,newNums[i] The assignment is (nums[i] + nums[i+1]) % 10 ,% Represents the remainder operation .
take newNums Replace array nums .
From step 1 Start repeating the whole process .
Please return nums Triangle and .

Example

Example 1：
Input ：nums = [1,2,3,4,5]
Output ：8
explain ：
The above figure shows the process of obtaining the triangular sum of arrays .

Example 2：
Input ：nums = [5]
Output ：5
explain ：
because nums There is only one element in , The triangle sum of the array is itself for this element .

The problem solving process

Ideas and steps

（1） The solution of double-layer circular violence , Or it can be solved by adding dynamic programming ;
（2） Pay attention to the inner layer when circulating  j  The cycle begins with  0  Start , To  nums.length - i  end , Because each cycle is one less number than the last time ;
（3） Finally, the last row of the array is returned  1  Just one element .

Code display

public class TriangularSum {
    

    public int triangularSum(int[] nums) {
    
        int[][] dp = new int[nums.length][nums.length];
        dp[0] = nums;
        for (int i = 1; i < nums.length; i++) {
    
            for (int j = 0; j < nums.length - i; j++) {
    
                dp[i][j] = (dp[i - 1][j] + dp[i - 1][j + 1]) % 10;
            }
        }
        //  Print 2D array 
        display(dp);
        return dp[nums.length - 1][0];
    }

    private void display(int[][] dp) {
    
        for (int i = 0; i < dp.length; i++) {
    
            for (int j = 0; j < dp[i].length; j++) {
    
                System.out.printf("%4d", dp[i][j]);
            }
            System.out.println();
        }
    }

    public static void main(String[] args) {
    
        int[] nums = {
    1, 2, 3, 4, 5};
        System.out.println(new TriangularSum().triangularSum(nums));
    }

}