119. Pascal's Triangle II

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

含义:要求计算杨辉三角第K行的元素,For example, given k = 3,Return [1,3,3,1].

思路:

对任意的n>0有 
  f(1, n)=1,(n>0) 
  f(n, n)=1,(n>2) 
  f(i,j) = f(i-1, j-1)+f(i, j-1),i>2,j>2, 
  求第k行。 

 1     public List<Integer> getRow(int rowIndex) {
 2      List<Integer> list = new ArrayList<Integer>();
 3     if (rowIndex < 0)
 4         return list;
 5 
 6     for (int i = 0; i < rowIndex + 1; i++) {
 7         list.add(0, 1);
 8         for (int j = 1; j < list.size() - 1; j++) {
 9             list.set(j, list.get(j) + list.get(j + 1));
10         }
11     }
12     return list;       
13     }
原文地址:https://www.cnblogs.com/wzj4858/p/7668555.html