时间轴

2025-12-13

init


题目:

注意要考虑 m==n==1 时的特殊情况。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
#include <vector>
using std::vector;

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid)
{
int i, j, m = obstacleGrid.size(), n = obstacleGrid[0].size();
if (m == 1 && n == 1) {
if (obstacleGrid[0][0] == 1) {
return 0;
}else{
return 1;
}
}
// dp[i][j]表示到达grid[i][j]的不同路径的数量
vector<vector<int> > dp(m, vector<int>(n, 0));

dp[0][0] = 1;
for (i = 1; i < m; i++) {
if (obstacleGrid[i - 1][0] != 1 &&
obstacleGrid[i][0] != 1) {
dp[i][0] = 1;
} else {
break;
}
}

for (j = 1; j < n; j++) {
if (obstacleGrid[0][j - 1] != 1 &&
obstacleGrid[0][j] != 1) {
dp[0][j] = 1;
} else {
break;
}
}

for (i = 1; i < m; i++) {
for (j = 1; j < n; j++) {
if (obstacleGrid[i][j] == 1) {
continue;
}
if (obstacleGrid[i - 1][j] == 1 &&
obstacleGrid[i][j - 1] == 1) {
continue; // set default as 0
} else if (obstacleGrid[i - 1][j] == 1 &&
obstacleGrid[i][j - 1] == 0) {
dp[i][j] = dp[i][j - 1];
} else if (obstacleGrid[i - 1][j] == 0 &&
obstacleGrid[i][j - 1] == 1) {
dp[i][j] = dp[i - 1][j];
} else {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
}
return dp[m - 1][n - 1];
}
};