【Leetcode】【Medium】Unique Paths II

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.

解题思路：

请参见Unique Paths I 的思路，对于Unique Paths II 我们依然希望使用o(n)的额外空间和o(m+n)的时间复杂度；

Unique Paths II中grid[i][n-1]和grid[i-1][n-1]不再总是相等，即格子中最右侧一列每一格存在的路径条数可能为1或0；

因此，为了继续使用Unique Paths I中数组迭代的技巧，需要每次迭代前计算新一轮的grid[i][n-1]值，这样才能继续计算grid[i][0]~grid[i][n-2]的值，从而完成此次迭代；

如果原始格子内为1，则对应此位值置0，表示从此位置不存在到终点的有效路径。

代码：

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
vector<int> col (n, 0);

col[n-1] = obstacleGrid[m-1][n-1] == 1 ? 0 : 1;
for (int i = m - 1; i >= 0; --i) {
col[n-1] = obstacleGrid[i][n-1] == 1 ? 0 : col[n-1];
for (int j = n - 2; j >= 0; --j) {
col[j] = obstacleGrid[i][j] == 1 ? 0 : col[j] + col[j+1];
}
}

return col[0];
}
};

注：

养成好习惯，除非特殊说明，不要在原始入参上修改，尤其是入参是地址形式。