【算法】【Dynamic Programming】Unique Paths II

Description

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.

Solution

思路

一开始的想法：创建 m*n 的矩阵 dp，

dp[i][j]

存储机器人到 (i, j) 的唯一路径数.

若该点无障碍：

dp[i][0] = dp[0][j] = 1, 0 <= i < m, 0 <= j < n

dp[i][j] = dp[i][j - 1] + dp[i - 1][j], 1 <= i < m, 0 <= j < n

若该点有障碍：

dp[i][j] = 0

后来在 Discuss 看到更好的答案, 它在上面实现的基础上节省了内存空间：创建长度为 n 的数组 dp, 进行 m 轮迭代. 每轮迭代中，

dp[i] += dp[i - 1]

, 相当于上面的

dp[i][j] = dp[i][j - 1] + dp[i - 1][j]

.

代码

class Solution {
public:
int uniquePathsWithObstacles(vector< vector< int > >& obstacleGrid) {
int row = obstacleGrid.size();
int col = obstacleGrid[0].size();
int dp[100];
memset(dp, 0, sizeof(int) * 100);
dp[0] = 1;

for (int r = 0; r < row; ++r) {
for (int c = 0; c < col; ++c) {
if (obstacleGrid[r][c] == 1) {
dp[c] = 0;
}
else {
if (c > 0)
dp[c] += dp[c - 1];
}
}
}
return dp[col - 1];
}
};