Triangle

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

For example, given the following triangle

[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]

The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).

Note:
Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

跟矩阵从左上到右下有多少种走法，以及最小和路径是一样的，朴素的DP需要O(n2)的空间，我们先写一个，其实如果要求O(n)的空间无非就是用滚动数组，也不难的。

#define vvi vector<vector<int> >
class Solution {
public:
    int minimumTotal(vvi& tri)
{
    vvi dp(tri);
	int n=tri.size();
	int i,j;
	for(i=1;i<n;i++)
	{
		int m=tri[i].size();
		dp[i][0]+=dp[i-1][0];
		for(j=1;j<m-1;j++)
			dp[i][j]=min(dp[i-1][j-1],dp[i-1][j])+tri[i][j];
		dp[i][m-1]+=dp[i-1][m-2];
	}
	return *min_element(dp[n-1].begin(),dp[n-1].end());
}


};

下面是用滚动数组的：

#define vvi vector<vector<int> >
class Solution {
public:
int minimumTotal(vvi& tri)
{
    int n=tri.size();
	if ( n==0 )
		return 0;

	vector<int> dp(1,tri[0][0]);
	int i,j;
	for(i=1;i<n;i++)
	{
		int m=tri[i-1].size();
		dp.push_back(dp[m-1]+tri[i][m]);
		for(j=m-1;j>0;j--)
			dp[j]=min(dp[j-1],dp[j])+tri[i][j];
		dp[0]+=tri[i][0];
	}
	return *min_element(dp.begin(),dp.end());
}



};