[Leetcode] Pascal's Triangle II

Given an index k, return the k^th row of the Pascal's triangle.

For example, given k = 3,
Return [1,3,3,1].

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

因为需要O（k）的空间，因此需要对同一个数组进行重写，可以从后往前进行覆盖，则没有问题

需要注意的一个问题是 Arrays.asList()操作的不能使基本类型数组

 1 public class Solution {
 2     public List<Integer> getRow(int rowIndex) {
 3         Integer [] row = new Integer[rowIndex+1];
 4         row[0]=1;
 5         if(rowIndex==0)return Arrays.asList(row);
 6         row[1]=1;
 7         if(rowIndex==1) return Arrays.asList(row);
 8         for(int i = 2; i<=rowIndex;i++){
 9             row[i]=1;
10             for(int j=i-1;j>=1;j--){
11                 row[j]=row[j]+row[j-1];
12             }
13         }
14         return Arrays.asList(row);
15     }
16 }

转载于:https://www.cnblogs.com/deepblueme/p/4736091.html

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