[LeetCode] Minimum Path Sum 解题报告

Given a m[/i] x n[/i] grid filled with non-negative numbers, find a path from top left to bottom right which minimizes[/i] the sum of all numbers along its path.Note:[/b] You can only move either down or right at any point in time.» Solve this problem

[解题报告]
二维DP。设数组A[row][col],
Min[i][j] = min(Min[i-1][j], Min[i][j-1]) +A[i][j];
注意初始条件即可。

[code]1:    int minPathSum(vector<vector<int> > &grid) {
2:      // Start typing your C/C++ solution below
3:      // DO NOT write int main() function
4:      if(grid.size() ==0) return 0;
5:      int row = grid.size();
6:      int col = grid[0].size();
7:      int Min[row][col];
8:      Min[0][0] =grid[0][0];
9:      for(int i =1; i < row; i ++)
10:      {
11:        Min[i][0] =Min[i-1][0] + grid[i][0];
12:      }
13:      for(int i =1; i< col; i++)
14:      {
15:        Min[0][i] = Min[0][i-1] + grid[0][i];
16:      }
17:      for(int i =1; i< row; i++)
18:      {
19:        for(int j =1; j< col; j++)
20:        {
21:          Min[i][j] = min(Min[i-1][j], Min[i][j-1]) + grid[i][j];
22:        }
23:      }
24:      return Min[row-1][col-1];
25:    }

Update: 3/16/2013. Refactor code
没必要用二维数组，用滚动数组即可。

1:    int minPathSum(vector<vector<int> > &grid) {
2:      int row = grid.size();
3:      if(row == 0) return 0;
4:      int col = grid[0].size();
5:      if(col == 0) return 0;
6:      vector<int> steps(col, INT_MAX);
7:      steps[0] =0;
8:      for(int i =0; i< row; i++)
9:      {
10:          steps[0] = steps[0] + grid[i][0];
11:          for(int j=1; j<col; j++)
12:          {
13:             steps[j]=min(steps[j], steps[j-1]) + grid[i][j];
14:          }
15:      }
16:      return steps[col-1];
17:    }

注意，Line 6，初始值是INT_MAX。因为Line 13里面用了min函数。