LeetCode: Unique Paths I & II & Minimum Path Sum

Title:

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

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.

思路：直观的思路是使用递归，但是会超时

class Solution{
public:
    int m;
    int n;
    int uniquePaths(int m, int n) {
        this->m = m;
        this->n = n;
        int sum = ;
        fun(,,sum);
        return sum;
    }
    void fun(int i,int j,int& sum){
        if (i == m && j == n)
            sum++;
        if (i > m || j > n)
            return ;
        fun(i+,j,sum);
        fun(i,j+,sum);
    }
};

int uniquePaths(int m,int n){
    if (m ==  || n == )
        return ;
    return uniquePaths(m-,n)+uniquePaths(m,n-);
}

一般这种递归都可以使用动态规划来解决

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

Unique Path II

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

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.

开始想直接使用I中的，却没有考虑到边界上有障碍的情况

int uniquePathsWithObstacles(vector<vector<int> > &obstacleGrid){
            if (obstacleGrid.empty())
                return ;
            int m = obstacleGrid.size();
            int n = obstacleGrid[].size();
            if (m <  || n < )
                return ;
            vector<int> result(n);
            result[] = ;
            for (int i =  ; i < m ; i++){
                for (int j =  ; j < n ; j++){
                    if (obstacleGrid[i][j] == )
                        result[j] = ;
                    else{
                        if (j > )
                            result[j] += result[j-];
                    }
                }
            }
            return result[n-];
        }

Minimun-Path-Sum

Title：

https://leetcode.com/problems/minimum-path-sum/

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.

思路：同样的动态规划

class Solution{
public:
    int minPathSum(vector<vector<int> > &grid){
        if (grid.empty() || grid.size() == )
            return ;
        int m = grid.size();
        int n = grid[].size();
        vector<int> v(n,INT_MAX);
        v[] = ;
        for (int i = ; i < m; i++){
            for (int j = ; j < n; j++){
                if (j == ){
                    v[j] = v[j] + grid[i][j];
                }else{
                    v[j] = min(v[j],v[j-]) + grid[i][j];
                }
                //cout<<v[j]<<" ";
            }
            //cout<<endl;
        }
        //cout<<endl;
        return v[n-];
    }
};