不同路径—动态规划

思路：第一行和第一列的元素都有一条路径可以到达，将对应位置赋值为1。接下来每个位置处都可以从其上边或左边到达。

 1 int uniquePaths(int m, int n) {
 2         // write your code here
 3         int ** s = new int*[m];
 4         for (int i = 0; i < m; i++) {
 5             s[i] = new int[n];
 6         }
 7         for (int i = 0; i < m; i++) {
 8             s[i][0] = 1;
 9         }
10         for (int i = 0; i < n; i++) {
11             s[0][i] = 1;
12         }
13         for (int i = 1; i < m; i++) {
14             for (int j = 1; j < n; j++) {
15                 s[i][j] = s[i-1][j] + s[i][j-1];
16             }
17         }
18         return s[m-1][n-1];
19     }

思路：第一行和第一列在没有遇到障碍时，都有一条路径可以到达，都设为1。但当第一行（列）某个位置遇到障碍时，该位置处和其右边（下边）位置处都不能到达了，都要设为0。接下来的位置，当遇到障碍时设为0，没有障碍时还是其上边的路径加上左边的路径。

 1 int uniquePathsWithObstacles(vector<vector<int>> &obstacleGrid) {
 2         // write your code here
 3         int row = obstacleGrid.size();
 4         int col = obstacleGrid[0].size();
 5         int ** s = new int*[row];
 6         for (int i = 0; i < row; i++) {
 7             s[i] = new int[col];
 8         }
 9         int t = 0, flag = 0;
10         for (int i = 0; i < row; i++) {
11             if (obstacleGrid[i][0] == 0) {
12                 s[i][0] = 1;
13             } else {
14                 s[i][0] = 0;
15                 t = i;
16                 flag = 1;
17                 break;
18             }
19         }
20         if (flag == 1) {
21             for (int i = t; i < row; i++) {
22                 s[i][0] = 0;
23             }
24             flag = 0;
25         }
26
27         for (int j = 0; j < col; j++) {
28             if (obstacleGrid[0][j] == 0) {
29                 s[0][j] = 1;
30             } else {
31                 s[0][j] = 0;
32                 t = j;
33                 flag = 1;
34                 break;
35             }
36         }
37         if (flag == 1) {
38             for (int j = t; j < col; j++) {
39                 s[0][j] = 0;
40             }
41             flag = 1;
42         }
43         for (int i = 1; i < row; i++) {
44             for (int j = 1; j < col; j++) {
45                 if (obstacleGrid[i][j] == 0) {
46                     s[i][j] = s[i-1][j] + s[i][j-1];
47                 } else {
48                     s[i][j] = 0;
49                 }
50             }
51         }
52         return s[row-1][col-1];
53     }