[leet code] Pascal's Triangle

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#algorithm #array list #Pascals triangle #leetcode #java

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本文详细介绍了如何使用迭代方式生成指定行数的Pascal's三角形，并提供了相应的代码实现。通过理解三角形的构造过程，读者可以轻松掌握生成三角形的方法。

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Given numRows, generate the first numRows of Pascal's triangle.

For example, given numRows = 5,
Return

[
     [1],
    [1,1],
   [1,2,1],
  [1,3,3,1],
 [1,4,6,4,1]
]

=============

Analysis:

The only difficulty for me is the definition of Pascal's triangle. After reading wikipedia:

Accordingly, we need only using 2 levels of iterations to build the Pascal's triangle. the first one is for each level/row in the triangle, and the second one is for each element in the current level/row.

public class Solution {
    public ArrayList<ArrayList<Integer>> generate(int numRows) {
        // result array list
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        
        // special case
        if (numRows == 0) return result;
        
        // construct the whole Pascal's triangle
        for (int i=1; i<=numRows; i++){
            // construct each level in Pascal's triangle
            ArrayList<Integer> curLevel = new ArrayList<Integer>();
            
            // level 1
            if (i==1) {
                curLevel.add(1);
                result.add(curLevel);
            }
            
            // other levels
            else{
                // get previous level of integer array list
                ArrayList<Integer> preLevel = result.get(i-2);
                
                // cunstruct current level of integer array list
                for (int j=1; j<=i; j++){
                    if(j==1 || j==i) curLevel.add(1);
                    else curLevel.add(preLevel.get(j-2) + preLevel.get(j-1));
                }
                result.add(curLevel);
            }
        }
        return result;
    }
}