Triangle <leetcode>

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.

 

算法：首先想到的是DFS，但是超时了代码如下：

class Solution {
public:
    int  result;
    int minimumTotal(vector<vector<int> > &triangle) {
        result=0;
        doit(triangle,0,0,0);
        return result;
    }
    void doit(vector<vector<int>>  &triangle,int level,int key,int sum)
    {
        if(level==triangle.size())
        {
           if(sum>result)  result=sum;   
        }
        if(level<triangle.size()-1)
        {
           doit(triangle,level+1,key,sum+triangle[level][key]);
           doit(triangle,level+1,key+1,sum+triangle[level][key]);
        }
    }
};

　　在网上查到的动态规划算法，从下到上，要求第i层到最底层路径的最小和，只要知道i+1层到最底层的最小路径和，代码如下：

class Solution {
public:
    int minimumTotal(vector<vector<int> > &triangle) {
    
        int len=triangle.size();
        if(len==0)  return -1;
        if(triangle[len-1].size()!=len)  return -1;
        int *temp=new int[len];
        for(int i=0;i<len;i++)
        {
            temp[i]=triangle[len-1][i];
        }
        for(int i=len-2;i>=0;i--)
        {
            for(int j=0;j<=i;j++)
            {
                temp[j]=triangle[i][j]+min(temp[j],temp[j+1]);
            }
        }
        return temp[0];
    }
   
};