之所以将这三道题放在一起，是因为这三道题非常类似。
1. Minimum Path Sum
题目要求：
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.
该题解答参考自一博文。
设dp[i][j]表示从左上角到grid[i][j]的最小路径和。那么dp[i][j] = grid[i][j] + min( dp[i-1][j], dp[i][j-1] );
下面的代码中，为了处理计算第一行和第一列的边界条件，我们令dp[i][j]表示从左上角到grid[i-1][j-1]的最小路径和，最后dp[rows][cols]是我们所求的结果
1 int minPathSum(vector<vector<int>>& grid) {
2 int rows = grid.size();
3 if(rows == 0)
4 return 0;
5 int cols = grid[0].size();
6
7 vector<vector<int> > dp(rows + 1, vector<int>(cols + 1, INT_MAX));
8 dp[0][1] = 0;
9 for(int i = 1; i < rows + 1; i++)
10 for(int j = 1; j < cols + 1; j++)
11 dp[i][j] = grid[i - 1][j - 1] + min(dp[i][j - 1], dp[i - 1][j]);
12
13 return dp[rows][cols];
14 }
注意到上面的代码中dp[i][j] 只和上一行的dp[i-1][j]和上一列的dp[i][j-1]有关，因此可以优化空间为O（n）（准确来讲空间复杂度可以是O（min（row,col）））
1 int minPathSum(vector<vector<int>>& grid) {
2 int rows = grid.size();
3 if(rows == 0)
4 return 0;
5 int cols = grid[0].size();
6
7 vector<int> dp(cols + 1, INT_MAX);
8 dp[1] = 0;
9 for(int i = 1; i < rows + 1; i++)
10 for(int j = 1; j < cols + 1; j++)
11 dp[j] = grid[i-1][j-1] + min(dp[j-1], dp[j]);
12
13 return dp[cols];
14 }
2. Unique Paths
题目要求：
A robot is located at the top-left corner of a m x n grid (marked ‘Start‘ in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish‘ in the diagram below).
How many possible unique paths are there?
Above is a 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
这道题的解答跟上一道题是非常类似的，程序如下：
1 int uniquePaths(int m, int n) {
2 if(m == 0 && n == 0)
3 return 0;
4
5 vector<vector<int> > dp(m, vector<int>(n, 1));
6 for(int i = 1; i < m; i++)
7 for(int j = 1; j < n; j++)
8 dp[i][j] = dp[i-1][j] + dp[i][j-1];
9
10 return dp[m-1][n-1];
11 }
优化空间程序：
1 int uniquePaths(int m, int n) {
2 if(m == 0 && n == 0)
3 return 0;
4
5 vector<int> dp(n, 1);
6 for(int i = 1; i < m; i++)
7 for(int j = 1; j < n; j++)
8 dp[j] = dp[j-1] + dp[j];
9
10 return dp[n - 1];
11 }
3. Unique Paths II
题目要求：
Follow up for "Unique Paths":
Now consider if some obstacles are added to the grids. How many unique paths would there be?
An obstacle and empty space is marked as 1 and 0 respectively in the grid.
For example,
There is one obstacle in the middle of a 3x3 grid as illustrated below.
[ [0,0,0], [0,1,0], [0,0,0] ]
The total number of unique paths is 2.
Note: m and n will be at most 100.
这道题跟上一道题基本一致，不同的地方在于我们需要将能到达存在obstacle的地方的路径数置为0。程序如下：
1 int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
2 int rows = obstacleGrid.size();
3 if(rows == 0)
4 return 0;
5
6 int cols = obstacleGrid[0].size();
7 if(cols == 0)
8 return 0;
9
10 vector<vector<int> > dp(rows, vector<int>(cols, 1));
11 int i = 0;
12 while(i < rows)
13 {
14 if(obstacleGrid[i][0] == 1)
15 while(i < rows)
16 {
17 dp[i][0] = 0;
18 i++;
19 }
20 i++;
21 }
22 int j = 0;
23 while(j < cols)
24 {
25 if(obstacleGrid[0][j] == 1)
26 while(j < cols)
27 {
28 dp[0][j] = 0;
29 j++;
30 }
31 j++;
32 }
33
34 for(i = 1; i < rows; i++)
35 for(j = 1; j < cols; j++)
36 {
37 if(obstacleGrid[i][j] == 1)
38 dp[i][j] = 0;
39 else
40 dp[i][j] = dp[i-1][j] + dp[i][j-1];
41 }
42
43 return dp[rows-1][cols-1];
44 }