LeetCode | Unique Paths I,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.

咋一看，深搜之，直接超时~~
然后发现是可以拿DP做的，思考出来一个DP方程，虽然不是很优化但是一次过了还是很开心的
以下代码展现了一些思考过程：
1、深搜
2、visit标记数组，仔细分析发现不会走过已经走的路
3、dp[i][j]=dp[i-1][j]+dp[i][j-1] 一开始想着倒推，然后一想似乎不能倒推，是从第一步开始走的

class Solution {
public:int dir[2][2]={{1,0},{0,1}};int route=0;int m,n;int uniquePaths(int m, int n) {this->m=m;this->n=n;// vector<vector<char> > visit(m,n);// char visit[m][n];// memset(visit,0,sizeof(visit));// dfs(1,1);// return route;// vector<vector<int> > dp(m,n);int dp[m+2][n+2];memset(dp,0,sizeof(dp));dp[1][1]=1;for(int i=1;i<=m;i++){for(int j=1;j<=n;j++){dp[i][j+1]=dp[i][j]+dp[i-1][j+1];dp[i+1][j]=dp[i][j]+dp[i+1][j-1];}}return dp[m][n];//不能倒推// for(int i=m-1;i>0;i--){//     for(int j=n-1;j>0;j--){//         dp[i][j]=dp[i-1][j]+dp[i][j-1];//     }// }}void dfs(int x,int y){if(x==m && y==n){route++;return;}for(int i=0;i<2;i++){int next_x=x+dir[i][0];int next_y=y+dir[i][1];//只存在越界，不存在反复访问if(next_x>m || next_y>n) continue;// if(visit[next_x][next_y])dfs(next_x,next_y);}}
};

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.

相比较I的话，II只是增加了一点障碍，使得在进入DP的时候需要一些条件。
由于不小心写错了一个条件所以导致debug好久

不过执行的时间似乎有点长，还有待优化：
参考链接：https://discuss.leetcode.com/topic/15267/4ms-o-n-dp-solution-in-c-with-explanations

class Solution {
public:int m,n;vector<vector<int> > obstacleGrid;int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {int m=obstacleGrid.size();if(m==0) return 0;int n=obstacleGrid[0].size();// if(m==1 && n==1) return !obstacleGrid[0][0];if(obstacleGrid[0][0]) return 0;int dp[m+2][n+2];this->m=m;this->n=n;this->obstacleGrid=obstacleGrid;memset(dp,0,sizeof(dp));dp[1][1]=1;for(int i=1;i<=m;i++){for(int j=1;j<=n;j++){if(!reachAble(i,j)) continue;//这一点是可达的if(reachAble(i,j+1)){dp[i][j+1]=dp[i][j]+reachAble(i-1,j+1)*dp[i-1][j+1];}if(reachAble(i+1,j)){dp[i+1][j]=dp[i][j]+reachAble(i+1,j-1)*dp[i+1][j-1];}}}return dp[m][n];}int reachAble(int x,int y){int i=x-1,j=y-1;if(i>=m || i<0 || j>=n || j<0) return 0;return !obstacleGrid[i][j];}
};

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