算法分析与设计——LeetCode Problem.63 Unique Paths II

题目链接

问题描述

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

这题与Leetcode 62解决方法相似，可以新建一个与grid相同size的矩阵path。需要注意的是grid的最左边一列grid[i][0]和最上边一排grid[0][i]：从grid的左上角grid[0][0]开始，分别沿最左边和最上边遍历，直到遍历完整一行/列或者遇到数字1，在此之前要把path相应位置赋值为1。之后根据递推式path[i][j] = path[i-1][j] + path[i][j-1]遍历path矩阵即可，若所遍历的位置path[i][j]在grid矩阵中的相应位置grid[i][j]的值为1，则将path[i][j]的值赋为0即可。

代码如下

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {
if(obstacleGrid.empty())
return 0;
int m = obstacleGrid.size();
if(obstacleGrid[0].empty())
return 0;
int n = obstacleGrid[0].size();
vector<vector<int> > path(m, vector<int>(n, 0));
for(int i = 0; i < m; i ++)
{
if(obstacleGrid[i][0] != 1)
path[i][0] = 1;

4000
else
break;
}
for(int i = 0; i < n; i ++)
{
if(obstacleGrid[0][i] != 1)
path[0][i] = 1;
else
break;
}
for(int i = 1; i < m; i ++)
{
for(int j = 1; j < n; j ++)
{
if(obstacleGrid[i][j] == 1)
path[i][j] = 0;
else
path[i][j] = path[i-1][j] + path[i][j-1];
}
}
return path[m-1][n-1];
}
};