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

Note: m and n will be at most 100.

Example 1:

Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right

Example 2:

Input: m = 7, n = 3
Output: 28

基本上就是裸的dp

class Solution {
public:
    int uniquePaths(int m, int n) {
        int num[109][109];
        for(int i=0;i<=m;i++)
            num[i][0]=1;
        for(int i=0;i<=n;i++)
            num[0][i]=1;
        for(int i=1;i<m;i++)
            for(int j=1;j<n;j++)
                num[i][j]=num[i-1][j]+num[i][j-1];
        return num[m-1][n-1];
    }
};

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).

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.

Note: m and n will be at most 100.

Example 1:

Input:
[
  [0,0,0],
  [0,1,0],
  [0,0,0]
]
Output: 2
Explanation:
There is one obstacle in the middle of the 3x3 grid above.
There are two ways to reach the bottom-right corner:
1. Right -> Right -> Down -> Down
2. Down -> Down -> Right -> Right

有障碍的dp 有障碍的地方变成0 注意边界处理

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
        int m=obstacleGrid.size();
        int n=obstacleGrid[0].size();
       // printf("m=%d,n=%d\n",m,n);
        int num[109][109];
        bool flag=0;
        for(int i=0;i<m;i++)
            if(obstacleGrid[i][0]==1||flag)
                num[i][0]=0,flag=1;
            else
                num[i][0]=1;
        flag=0;
        for(int i=0;i<n;i++)
            if(obstacleGrid[0][i]==1||flag)
                num[0][i]=0,flag=1;
            else
                num[0][i]=1;
        for(int i=1;i<m;i++)
            for(int j=1;j<n;j++)
                if(obstacleGrid[i][j]==0)
                    num[i][j]=num[i-1][j]+num[i][j-1];
                else
                    num[i][j]=0;
        return num[m-1][n-1];

    }
};

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

Example:

Input:
[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]
Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

求路径最小和

（侮辱智商

class Solution {
public:
    int minPathSum(vector<vector<int>>& grid) {
        int dp[1000][1000];
        int m=grid.size();
        int n=grid[0].size();
        dp[0][0]=grid[0][0];
        for(int i=1;i<m;i++)
            dp[i][0]=grid[i][0]+dp[i-1][0];
        for(int i=1;i<n;i++)
            dp[0][i]=grid[0][i]+dp[0][i-1];
        for(int i=1;i<m;i++)
            for(int j=1;j<n;j++)
                dp[i][j]=min(dp[i-1][j],dp[i][j-1])+grid[i][j];
        return dp[m-1][n-1];
    }
};

这几个题给我一种leetcode 中等题不过如此的感觉