标签：style blog color io ar strong for sp div
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];
}
};
标签：style blog color io ar strong for sp div
原文地址：http://www.cnblogs.com/sqxw/p/4001010.html
