LeetCode 119. Pascal's Triangle II

分析

难度 易

来源

https://leetcode.com/problems/pascals-triangle-ii/

题目

Given a non-negative index k where k ≤ 33, return the k^th index row of the Pascal's triangle.

Note that the row index starts from 0.

 

In Pascal's triangle, each number is the sum of the two numbers directly above it.

Example:

Input: 3

Output: [1,3,3,1]

Follow up:

Could you optimize your algorithm to use only O(k) extra space?

解答

package LeetCode;

import java.util.ArrayList;
import java.util.List;

public class L119_PascalTriangleII {
    public List<Integer> getRow(int rowIndex) {
        if(rowIndex<0)
            return null;
        //从0开始的第k行，即（k+1）对应斐波那契数列的最后一行
        int listLen=rowIndex+1;
        List<Integer> list=new ArrayList<Integer>();//记录当前层数值
        for(int i=0;i<listLen;i++)//不知道为什么这句写在这里比下边二层循环的外层快，
            list.add(1);//整个都赋1，初试值，各层边界值       
        for(int i=0;i<listLen;i++){
            for(int j=i-1;j>0;j--){//一行一行赋值，对每行从右向左赋值，避免修改下一个数字要用到两个加数
                list.set(j,list.get(j-1)+list.get(j));
            }
        }
        return list;
    }
    public static void main(String[] args){
        long time1=System.currentTimeMillis();
        L119_PascalTriangleII l119=new L119_PascalTriangleII();
        System.out.println(l119.getRow(10));//从0开始的第k行，即（k+1）对应斐波那契数列的最后一行
        long time2=System.currentTimeMillis();
        System.out.println(time2-time1);
    }
}