leetcode: Pascal's Triangle II

iven 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?

要求只能用O(k)空间，所以用(rowIndex+1)*2的数组来存储结果，每次交换cur和pre。

状态转换方程dp[cur][i] = dp[pre][i-1] + dp[pre][i];

class Solution {public:    vector<int> getRow(int rowIndex) {        vector< vector< int> > dp( 2, vector< int>( rowIndex+1, 0));        int cur = 1;        int pre = 0;        dp[pre][0] = 1;        dp[cur][0] = 1;        int n = 1;        while( n <= rowIndex){            for( int i = 1; i < n + 1; ++i){                dp[cur][i] = dp[pre][i-1] + dp[pre][i];            }            ++n;            cur = !cur;            pre = !pre;        }        return dp[pre];    }};