Minimum Path Sum Total
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.
Note: You can only move either down or right at any point in time.
解题思路:遍历数组,对于每一个点,要么从左边走过来,要么从上面
走过来,选择路径最短的一条。
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.
Note: You can only move either down or right at any point in time.
解题思路:遍历数组,对于每一个点,要么从左边走过来,要么从上面
走过来,选择路径最短的一条。
#include<iostream>
#include<vector>
using namespace std;
class Solution {
public:
int minPathSum(vector<vector<int> > &grid) {
int m = grid.size();
int n = grid[0].size();
for(int i=1;i<n;++i)
grid[0][i]+=grid[0][i-1];
for(int i=1;i<m;++i)
grid[i][0]+=grid[i-1][0];
for(int i=1;i<m;++i){
for(int j=1;j<n;++j){
grid[i][j]+=(grid[i-1][j]>grid[i][j-1]?grid[i][j-1]:grid[i-1][j]);
}
}
return grid[m-1][n-1];
}
};
int main(){
vector< vector<int> > grid;
vector<int> vint0;
vint0.push_back(1);
/*vint0.push_back(2);
vint0.push_back(3);
vint0.push_back(4);
grid.push_back(vint0);
grid.push_back(vint0);
grid.push_back(vint0);
grid.push_back(vint0);*/
grid.push_back(vint0);
Solution s;
cout<<s.minPathSum(grid)<<endl;
return 0;
}