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因为被网上的另外一种算法吸引了。

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];
    }
};

第二种是采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)步而其中我们可以选择(m-1)+(n-1)任意的（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

这种解法的缺点是可能在m，n取较大的数值时候无法储存。所以此处我们采用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.

这里只是加了障碍物而已，在障碍物的位子是0，

还有初始化只能初始化第一个位子即起点，如果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];
    }
};