LeetCode #118 Pascal’s Triangle Solution & Explanation

LeetCode Problem

Given an integer numRows, return the first numRows of Pascal’s triangle.

In Pascal’s triangle, each number is the sum of the two numbers directly above it as shown:

This question is similar to LeetCode #70 – Climbing Stairs (LeetCode #70 Climbing Stairs Solution & Explanation – zyrastory Code & Food Research Center)

Solution

C# Solution

public class Solution {
    public IList<IList<int>> Generate(int numRows) {
        List<List<int>>res = new List<List<int>>();
        
        
        res.Add(new List<int>(){1});   //the first line
        
        if(numRows!=1)  
        {
            for(int i = 1;i<numRows;i++)
            {
                List<int>tmp = new List<int>();  //temp array
                for(int j=0;j<=i;j++)
                {
                    int num;
                    if(j==0 || j==i)   //the first and end number of a line is one
                    {
                        num = 1;
                    }
                    else
                    {
                        num = res[i-1][j-1]+res[i-1][j];
                    }
                    tmp.Add(num); 
                }
                res.Add(tmp);  //add temp array to the result
            }
        }
        
        return res.ToArray();
        
    }
}

Java Solution

class Solution {
    public List<List<Integer>> generate(int numRows) {
        
        List<List<Integer>>res = new ArrayList<List<Integer>>();
        
        res.add(Arrays.asList(new Integer[]{1}));
        
        if(numRows!=1)
        {
            for(int i = 1;i<numRows;i++)
            {
                List<Integer>tmp = new ArrayList<Integer>();
                for(int j=0;j<=i;j++)
                {
                    int num;
                    if(j==0 || j==i)
                    {
                        num = 1;
                    }
                    else
                    {
                        num = res.get(i-1).get(j-1)+res.get(i-1).get(j);
                    }
                    tmp.add(num);
                }
                res.add(tmp);
            }
        }
        
        return res;
    }
}

Runtime : 1ms

Python3 Solution

class Solution:
    def generate(self, numRows: int) -> List[List[int]]:
        
        res = [];
        res.append([1]);
        
        if numRows!=1:
            for i in range(1,numRows):
                tmp = [];
                for j in range(0,i+1):
                    if j==0 or j==i :
                        tmp.append(1);
                    else:
                        tmp.append(res[i-1][j-1]+res[i-1][j]);
                        
                res.append(tmp);
                
        return res;

JavaScript Solution

Solution1

/**
 * @param {number} numRows
 * @return {number[][]}
 */
var generate = function(numRows) {
    var res = [];
    res[0] = [1];
    
    if(numRows!=1)
    {
        for(let i = 1;i<numRows;i++)
        {
            let tmp = [];
            for(let j=0;j<=i;j++)
            {
                var num;
                if(j==0 || j==i)
                {
                    num = 1;
                }
                else
                {
                    num = res[i-1][j-1]+res[i-1][j];
                }
                tmp[j] = num;
            }
            res[i] = tmp;
        }
    }
    return res;
};

Conclusion

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The problem link : Pascal’s Triangle – LeetCode

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