Unique Paths I,II

题目来自于：https://leetcode.com/problems/unique-paths/

：https://leetcode.com/problems/unique-paths-ii/

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

How many possible unique paths are there?



Above is a 3 x 7 grid. How many possible unique paths are there?

Note: m and n will be at most 100.

这道题目就是典型的动态规划问题。之所以会写博客也是由于被网上的第二种算法吸引了。

典型的解法记住空间复杂度要在O（n）

class Solution {
public:
int uniquePaths(int m, int n) {
vector<int> paths(n,1);
for(int i=1;i<m;i++)
for(int j=1;j<n;j++)
paths[j]+=paths[j-1];
return paths[n-1];
}
};

另外一种是採用排列组合的方法来解答的

我们从左上角走到右下角一共要(m-1)+(n-1)步而当中我们能够选择(m-1)+(n-1)随意的（m-1）步向右，或者是（n-1）步向下。所以问题的答案就是Ian单的



这样的解法的缺点是可能在m。n取较大的数值时候无法储存。所以此处我们採用long int,

class Solution {
public:
int uniquePaths(int m, int n) {// (m-1 + n-1)! / ((m-1)! * (n-1)!)
int large = max(m,n) -1;
int small = min(m,n) -1;
if (large == 0 || small == 0) return 1;
long int numerator = 1, denominator = 1;
for (int i=1; i<=small; ++i){
numerator *= large + i;
denominator *= i;
}
return numerator/denominator;
}
};

Unique Paths II

Total Accepted: 35700 Total
Submissions: 127653My Submissions

Question
Solution

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.

这里仅仅是加了障碍物而已。在障碍物的位子是0，

还有初始化仅仅能初始化第一个位子即起点。假设起点不是障碍物则为1，否则是0；

class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
vector<int> paths(obstacleGrid[0].size(),0);
paths[0]=!obstacleGrid[0][0];
for(int i=0;i<obstacleGrid.size();++i)
for(int j=0;j<obstacleGrid[0].size();++j)
if(obstacleGrid[i][j]==1)
paths[j]=0;
else if(j-1>=0)
paths[j]+=paths[j-1];
return paths[obstacleGrid[0].size()-1];
}
};