Two-dimensional dp written by myself
class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
if(grid.empty())
return 0;
int m = grid.size();
int n = grid[0].size();
vector<vector<int>> dp(m,vector<int>(n));
dp[0][0] = grid[0][0];
for(int j = 1;j < n;j++){
dp[0][j] = grid[0][j] + dp[0][j-1];
}
for(int i = 1;i < m;i++){
dp[i][0] = grid[i][0] + dp[i-1][0];
}
for(int i = 1;i < m;i++){
for(int j = 1;j < n;j++){
dp[i][j] = min(dp[i-1][j],dp[i][j-1]) + grid[i][j];
}
}
return dp[m-1][n-1];
}
};
compressed space
class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
if(grid.empty())
return 0;
int m = grid.size();
int n = grid[0].size();
vector<int> dp(n);
for(int i = 0;i < m;i++){
for(int j = 0;j < n;j++){
if(i == 0 && j == 0){
dp[0] = grid[0][0];
//不加这个会溢出的
}else if(i == 0 && j > 0){
dp[j] = dp[j-1] + grid[i][j];
}else if(j == 0 && i > 0){
dp[j] = dp[j] + grid[i][j];
}else{
dp[j] = min(dp[j],dp[j-1]) + grid[i][j];
}
}
}
return dp[n-1];
}
}
The key to compressing the space is that currently in row i and column j, the number on column j-1 goes right to the j-th position, and the j-th position goes from the top to the j-th position.
If you want to understand, you can compress.