Given an index k, return the kth 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?
为了达到O(k)的空间复杂度要求，那么就要从右向左生成结果。相当于你提前把上一行的计算出来，当前行就可以用上一次计算出的结果计算了。
Java code:
public List<Integer> getRow(int rowIndex) { List<Integer> result = new ArrayList<Integer>(); if(rowIndex < 0) { return result; } result.add(1); for(int i = 1; i <= rowIndex; i++) { for(int j = result.size()-2; j >= 0; j--) { //from right to left result.set(j+1, result.get(j) + result.get(j+1)); } result.add(1); } return result; }
Reference:
1. http://www.programcreek.com/2014/04/leetcode-pascals-triangle-ii-java/
原文：http://www.cnblogs.com/anne-vista/p/4856637.html