Link: https://leetcode.cn/problems/find-triangular-sum-of-an-array/

Question

difficulty: mid
adj diff: 3

You are given a 0-indexed integer array nums, where nums[i] is a digit between 0 and 9 (inclusive).

The triangular sum of nums is the value of the only element present in nums after the following process terminates:

Let nums comprise of n elements. If n == 1, end the process. Otherwise, create a new 0-indexed integer array newNums of length n - 1.
For each index i, where 0 <= i < n - 1, assign the value of newNums[i] as (nums[i] + nums[i+1]) % 10, where % denotes modulo operator.
Replace the array nums with newNums.
Repeat the entire process starting from step 1.
Return the triangular sum of nums.

 

Example 1:


Input: nums = [1,2,3,4,5]
Output: 8
Explanation:
The above diagram depicts the process from which we obtain the triangular sum of the array.
Example 2:

Input: nums = [5]
Output: 5
Explanation:
Since there is only one element in nums, the triangular sum is the value of that element itself.
 

Constraints:

1 <= nums.length <= 1000
0 <= nums[i] <= 9

并没有什么牛逼的数学算法。

直接实现就行！

（我曾经尝试着一个牛逼的 mathematical 算法，结果失败了。。。代码如下）

class Solution {
    public int triangularSum(int[] nums) {
        // for 1 levels: 1*nums[0]
        // for 2 levels: 1*nums[0] + 1*nums[1]
        // for 3 levels: 1*nums[0] + 2*nums[1] + 1*nums[2]
        // for 4 levels: 1*nums[0] + 3*nums[1] + 3*nums[2] + 1*nums[3]
        // for 5 levels: 1*nums[0] + 4*nums[1] + 6*nums[2] + 4*nums[3] + 1*nums[4]
        // for 6 levels: 1*nums[0] + 5*nums[1] + 10*nums[2] + 10*nums[3] + 5*nums[4] + 1*nums[5]
        // if i = level
        int len = nums.length;
        long[] times = new long[len];
        times[0] = 1;
        for (int i = 1; i < len; i++) {
            for (int j = i; j >= 1; j--) {
                times[j] = times[j - 1] + times[j];
            }
        }

        long sum = 0;
        for (int i = 0; i < len; i++) {
            sum += (long) nums[i] * times[i];
            sum %= 10l;
        }
        return (int) sum;
    }
}

以上代码 170 / 300 个通过测试用例。

Code

直接 - 杨辉三角变式。

class Solution {
    public int triangularSum(int[] nums) {
        int len = nums.length;
        for (int i = len - 1; i >= 0; i--) {
            for (int j = 0; j < i; j++) {
                nums[j] = (nums[j + 1] + nums[j]) % 10;
            }
        }
        return nums[0];
    }
}

2221. Find Triangular Sum of an Array

Question

Code