[leetcode]120. Triangle

题目链接：https://leetcode.com/problems/triangle/#/description

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).

方法一：（自底向上）
class Solution {
public:
int minimumTotal(vector<vector<int> > &triangle)
{
vector<int> mini = triangle[triangle.size()-1];
for ( int i = triangle.size() - 2; i>= 0 ; --i )
for ( int j = 0; j < triangle[i].size() ; ++ j )
mini[j] = triangle[i][j] + min(mini[j],mini[j+1]);
return mini[0];
}
};

方法二：（自顶向下）
class Solution{
public:
int minimumTotal(vector<vector<int>>& triangle)
{

vector<vector<int>> minilen=vector<vector<int>>(triangle.size());
for(int i=0;i<triangle.size();i++)
minilen[i].resize(triangle[i].size());
minilen[0][0]=triangle[0][0];
for(int i=1;i<triangle.size();i++)
{
for(int j=0;j<triangle[i].size();j++)
{
if(j==0)
minilen[i][j]=minilen[i-1][j]+triangle[i][j];
else if(j==triangle[i].size()-1)
minilen[i][j]=minilen[i-1][j-1]+triangle[i][j];
else
minilen[i][j]=min(minilen[i-1][j-1],minilen[i-1][j])+triangle[i][j];
}
}
int minres=INT32_MAX;
for(int i=0;i<minilen.back().size();i++)
{
if (minilen.back()[i]<minres)
minres=minilen.back()[i];
}
return minres;
}
};