这道题用动态规划的思路解决。对于某一点(m,n)的不同路径数是(m-1,n)(m,n-1)两点的不同路径和(障碍物为0)。首先,我们初始化最上和最左边界上的点,然后遍历内部点。
注意特殊情况,如只有一行/一列,起始点为障碍物。
时间复杂度:O(N*M)
C++代码:
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int row = obstacleGrid.size(), col = obstacleGrid[0].size();
if (obstacleGrid[0][0] == 1)
return 0;
if (row == 1)
{
for (auto x : obstacleGrid[0])
if (x == 1)
return 0;
return 1;
}
if (col == 1)
{
for (int i = 0; i < row; i++)
if (obstacleGrid[i][0] == 1)
return 0;
return 1;
}
vector<vector<int>> record(row);
for (auto &x : record)
x.resize(col);
record[0][0] = 1;
for (int i = 1; i < col; i++)
(obstacleGrid[0][i] != 1) ? record[0][i] = record[0][i - 1] : 0;
for (int i = 1; i < row; i++)
(obstacleGrid[i][0] != 1) ? record[i][0] = record[i - 1][0] : 0;
for (int i = 1; i < row; i++)
for (int j = 1; j < col; j++)
(obstacleGrid[i][j] != 1) ? record[i][j] = record[i - 1][j] + record[i][j - 1] : 0;
return record[row - 1][col - 1];
}
};