问题描述：

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?

思路：

the mth element of the nth row of the Pascal‘s triangle is C(n, m) = n!/(m! * (n-m)!)

C(n, m-1) = n!/((m-1)! * (n-m+1)!)

so C(n, m) = C(n, m-1) * (n-m+1) / m

In additional, C(n, m) == C(n, n-m)

代码：

public List<Integer> getRow(int rowIndex) {
		if(rowIndex < 0)
			return new ArrayList<Integer>();
		
		
		int num = rowIndex+1;
		List<Integer> list = new ArrayList<Integer>(num);
		
	
		double [] factor = new double[num];
		double result = 1;
		factor[0] = 1;
		list.add(1);
		for(int i=1; i<num ; i++){
			result = result*(num-i)/i;
			factor[i] = result;
			list.add((int)factor[i]);
		}
		return list;
        
        
    }

[leetcode]Pascal's Triangle II