LeetCode | Pascal's Triangle

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]
]
//杨辉三角:每一行,头尾为1,中间元素为其上一行两肩位置元素之和
//第i行row的长度为i
public class Solution {
    public List<List<Integer>> generate(int numRows) {
        
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if(numRows < 1) return result;
        
        List<Integer> preRow = new ArrayList<Integer>();
        preRow.add(0);                       //表示上一行Row
        
        for(int i=1; i<=numRows; i++){       //每一行Row的长度就为i
            List<Integer> curRow = new ArrayList<Integer>();
            for(int j=0; j<i; j++){          //填充当前行
                if(j < 1){                      //head
                    curRow.add(1);
                }else if(j >= preRow.size()){   //tail
                    curRow.add(1);
                }else{
                    curRow.add(preRow.get(j-1) + preRow.get(j));
                }
            }
            
            preRow = curRow;
            result.add(curRow);
        }
        
        return result;
    }
}



原文地址:https://www.cnblogs.com/dosmile/p/6444437.html