文章目录
所有的LeetCode题解索引,可以看这篇文章------【算法和数据结构】LeetCode题解。
一、题目
二、解法
思路分析:参考【算法与数据结构】62、LeetCode不同路径的题目,可以发现本题仅仅是多了障碍物。我们还是用动态规划来做。有障碍物的地方无法到达,因此路径数量为0,只需要将障碍物位置的dp数组记为0,除此之外障碍物后面的位置有可能无法到达(程序当中的两个if break语句)。
程序如下:
cpp
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int row = obstacleGrid.size(), col = obstacleGrid[0].size();
vector<vector<int>> dp(row, vector<int>(col, 0));
for (int i = 0; i < col; i++) {
if (obstacleGrid[0][i] == 1) break;
dp[0][i] = 1;
}
for (int j = 0; j < row; j++) {
if (obstacleGrid[j][0] == 1) break;
dp[j][0] = 1;
}
for (int i = 1; i < row; i++) {
for (int j = 1; j < col; j++) {
if (obstacleGrid[i][j] == 0) dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
else dp[i][j] = 0;
}
}
return dp[row - 1][col - 1];
}
};
复杂度分析:
- 时间复杂度: O ( r o w ∗ c o l ) O(row*col) O(row∗col),,row和col分别是地图的行和列。
- 空间复杂度: O ( r o w ∗ c o l ) O(row*col) O(row∗col)。
三、完整代码
cpp
# include <iostream>
# include <vector>
using namespace std;
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int row = obstacleGrid.size(), col = obstacleGrid[0].size();
vector<vector<int>> dp(row, vector<int>(col, 0));
for (int i = 0; i < col; i++) {
if (obstacleGrid[0][i] == 1) break;
dp[0][i] = 1;
}
for (int j = 0; j < row; j++) {
if (obstacleGrid[j][0] == 1) break;
dp[j][0] = 1;
}
for (int i = 1; i < row; i++) {
for (int j = 1; j < col; j++) {
if (obstacleGrid[i][j] == 0) dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
else dp[i][j] = 0;
}
}
return dp[row - 1][col - 1];
}
};
int main() {
//vector<vector<int>> obstacleGrid = { {0, 0, 0}, {0, 1, 0}, { 0, 0, 0 } };
vector<vector<int>> obstacleGrid = { {0, 1}, {0, 0} };
Solution s1;
int result = s1.uniquePathsWithObstacles(obstacleGrid);
cout << result << endl;
system("pause");
return 0;
}
end