一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为 “Start” ）。
机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为 “Finish”）。
现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径？
网格中的障碍物和空位置分别用 1 和 0 来表示。
示例 1：
输入：obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]]
输出：2
解释：3x3 网格的正中间有一个障碍物。
从左上角到右下角一共有 2 条不同的路径：
1. 向右 -> 向右 -> 向下 -> 向下
2. 向下 -> 向下 -> 向右 -> 向右
示例 2：
输入：obstacleGrid = [[0,1],[0,0]] 输出：1
提示：
m == obstacleGrid.lengthn == obstacleGrid[i].length1 <= m, n <= 100obstacleGrid[i][j]为0或1
动态规划。
假设 dp[i][j] 表示到达网格 (i,j) 的路径数，先初始化 dp 第一列和第一行的所有值，然后判断。
- 若
obstacleGrid[i][j] == 1，说明路径数为 0，dp[i][j] = 0； - 若
obstacleGrid[i][j] == 0，则dp[i][j] = dp[i - 1][j] + dp[i][j - 1]。
最后返回 dp[m - 1][n - 1] 即可。
class Solution:
def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
m, n = len(obstacleGrid), len(obstacleGrid[0])
dp = [[0] * n for _ in range(m)]
for i in range(m):
if obstacleGrid[i][0] == 1:
break
dp[i][0] = 1
for j in range(n):
if obstacleGrid[0][j] == 1:
break
dp[0][j] = 1
for i in range(1, m):
for j in range(1, n):
if obstacleGrid[i][j] == 0:
dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
return dp[-1][-1]class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
int m = obstacleGrid.length, n = obstacleGrid[0].length;
int[][] dp = new int[m][n];
for (int i = 0; i < m && obstacleGrid[i][0] == 0; ++i) {
dp[i][0] = 1;
}
for (int j = 0; j < n && obstacleGrid[0][j] == 0; ++j) {
dp[0][j] = 1;
}
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
if (obstacleGrid[i][j] == 0) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
}
return dp[m - 1][n - 1];
}
}class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size(), n = obstacleGrid[0].size();
vector<vector<int>> dp(m, vector<int>(n));
for (int i = 0; i < m && obstacleGrid[i][0] == 0; ++i) {
dp[i][0] = 1;
}
for (int j = 0; j < n && obstacleGrid[0][j] == 0; ++j) {
dp[0][j] = 1;
}
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
if (obstacleGrid[i][j] == 0) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
}
return dp[m - 1][n - 1];
}
};func uniquePathsWithObstacles(obstacleGrid [][]int) int {
m, n := len(obstacleGrid), len(obstacleGrid[0])
dp := make([][]int, m)
for i := 0; i < m; i++ {
dp[i] = make([]int, n)
}
for i := 0; i < m && obstacleGrid[i][0] == 0; i++ {
dp[i][0] = 1
}
for j := 0; j < n && obstacleGrid[0][j] == 0; j++ {
dp[0][j] = 1
}
for i := 1; i < m; i++ {
for j := 1; j < n; j++ {
if obstacleGrid[i][j] == 0 {
dp[i][j] = dp[i-1][j] + dp[i][j-1]
}
}
}
return dp[m-1][n-1]
}function uniquePathsWithObstacles(obstacleGrid: number[][]): number {
const m = obstacleGrid.length;
const n = obstacleGrid[0].length;
const dp = Array.from({ length: m }, () => new Array(n).fill(0));
for (let i = 0; i < m; i++) {
if (obstacleGrid[i][0] === 1) {
break;
}
dp[i][0] = 1;
}
for (let i = 0; i < n; i++) {
if (obstacleGrid[0][i] === 1) {
break;
}
dp[0][i] = 1;
}
for (let i = 1; i < m; i++) {
for (let j = 1; j < n; j++) {
if (obstacleGrid[i][j] === 1) {
continue;
}
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
return dp[m - 1][n - 1];
}impl Solution {
pub fn unique_paths_with_obstacles(obstacle_grid: Vec<Vec<i32>>) -> i32 {
let m = obstacle_grid.len();
let n = obstacle_grid[0].len();
if obstacle_grid[0][0] == 1 || obstacle_grid[m - 1][n - 1] == 1 {
return 0;
}
let mut dp = vec![vec![0; n]; m];
for i in 0..n {
if obstacle_grid[0][i] == 1 {
break;
}
dp[0][i] = 1;
}
for i in 0..m {
if obstacle_grid[i][0] == 1 {
break;
}
dp[i][0] = 1;
}
for i in 1..m {
for j in 1..n {
if obstacle_grid[i][j] == 1 {
continue;
}
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
dp[m - 1][n - 1]
}
}