Leetcode-64_. Minimum Path Sum(最小路径和)—动态规划解法+记忆化搜索解法-C++解

Leetcode 64. Minimum Path Sum(最小路径和)

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

[[1,3,1],
[1,5,1],
[4,2,1]]

1→3→1→1→1

这道题也是典型的动态规划问题，但是我一开始用动态规划做得有点费劲，就用记忆化搜索先做出来了：

int m,n,res;
vector<vector<int> > memo;
int minPath(int i,int j,vector<vector<int> >& grid)
{
if(i==m-1&&j==n-1)
return grid[i][j];
if(memo[i][j]!=-1)
return memo[i][j];
if(i==m-1)
{
return minPath(i,j+1,grid)+grid[i][j];
}
if(j==n-1)
{
return minPath(i+1,j,grid)+grid[i][j];
}
res=min(minPath(i,j+1,grid),minPath(i+1,j,grid))+grid[i][j];
memo[i][j] = res;
return res;
}
int minPathSum(vector<vector<int> >& grid) {
m=grid.size();//行
n=grid[0].size();//列
memo=vector<vector<int> >(m,vector<int> (n,-1));
minPath(0,0,grid);
}

然后后面看了别人的代码才把动态规划的部分理解了：

class Solution {
public:

int minPathSum(vector<vector<int>>& grid) {
int m = grid.size();
int n = grid[0].size();

vector<vector<int> > sum(m, vector<int>(n, grid[0][0]));

for (int i = 1; i < m; i++)
sum[i][0] = sum[i - 1][0] + grid[i][0];
for (int j = 1; j < n; j++)
sum[0][j] = sum[0][j - 1] + grid[0][j];
for (int i = 1; i < m; i++)
for (int j = 1; j < n; j++)
sum[i][j]  = min(sum[i - 1][j], sum[i][j - 1]) + grid[i][j];
return sum[m - 1][n - 1];

}
};

自制测试用例

#include <iostream>
#include<vector>
using namespace std;

int m,n,res;
vector<vector<int> > memo;
int minPathSum(vector<vector<int> >& grid) {
m=grid.size();//行
n=grid[0].size();//列
memo=vector<vector<int> >(m,vector<int> (n,grid[0][0]));
for (int i = 1; i < m; i++)
memo[i][0] = memo[i - 1][0] + grid[i][0];
for (int j = 1; j < n; j++)
memo[0][j] = memo[0][j - 1] + grid[0][j];

for (int i = 1; i < m; i++)
for (int j = 1; j < n; j++)
memo[i][j]  = min(memo[i - 1][j], memo[i][j - 1]) + grid[i][j];
return memo[m - 1][n - 1];
}

int main(int argc, char *argv[]) {
vector<vector<int> >tt ;
vector<int> v;
v.push_back(4);
v.push_back(3);
v.push_back(1);
tt.push_back(v);
v.clear();
v.push_back(6);
v.push_back(2);
v.push_back(3);
tt.push_back(v);
v.clear();
v.push_back(4);
v.push_back(2);
v.push_back(1);
tt.push_back(v);
v.clear();
for(int i=0;i<tt.size();i++)

b85e
{
for(int j=0;j<tt[0].size();j++)
cout<<tt[i][j];
cout<<endl;
}
int min = minPathSum(tt);
cout<<"min:"<<min;
return 0;
}