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?
class Solution { public: vector<int> getRow(int rowIndex) { vector<int> rst; if(rowIndex<0) return rst; rst.push_back(1); for(int i=1;i<=rowIndex;i++){ rst.push_back(1); for (int j = i/2; j >0; j--){ rst[j]=rst[j]+rst[j-1]; } for (int m =1,n=i-1;n>m;--n,++m){ rst[n]=rst[m]; } } return rst; } };
LeetCode之Pascal's Triangle II,布布扣,bubuko.com
原文:http://blog.csdn.net/smileteo/article/details/23451495