到某一点的最小路径是这一点上面一点的最小路径与这一点左面一点的最小路径,二者的最小值,即D[n][m]=min{D[n-1][m],D[n][m-1]}。首先,要初始化最左边和最上边的边界,然后再对内部点进行遍历。
时间复杂度:O(M*N)
C++代码:
class Solution {
public:
int minPathSum(vector<vector<int>>& grid) {
int row = grid.size(), col = grid[0].size();
if (row == 1 && col == 1)
return grid[0][0];
vector<vector<int>> record(row);
for (auto &x : record)
x.resize(col);
record[0][0] = grid[0][0];
for (int i = 0; i < row - 1; i++)
{
record[i + 1][0] += record[i][0] + grid[i+1][0];
}
for (int i = 0; i < col - 1; i++)
{
record[0][i + 1] += record[0][i] + grid[0][i + 1];
}
for(int i = 1;i<row;i++)
for (int j = 1; j < col; j++)
{
record[i][j] = min(record[i - 1][j], record[i][j - 1]) + grid[i][j];
}
return record[row - 1][col - 1];
}
};