[Leetcode]Unique Paths

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.

一种使用数学方法,简单排列组合。注意可能超过int，使用double，最后强制类型转换。

class Solution {
public:
    int uniquePaths(int m, int n) {
        if(m <=0 || n <= 0) return 0;
        long long res = 1;
        for(int i = n; i < m+n-1 ; i++){
            res = res * i / (i- n + 1);
        }
        return (int)res;
    }
};

　　

另外一种dp。

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

　

Unique Paths II

Total Accepted: 48928 Total Submissions: 173558 Difficulty: Medium

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.

dp做了一些改变。

class Solution {
public:
    int uniquePathsWithObstacles(vector<vector<int>>& g) {
        int m = g.size();
        int n =g[0].size();
        vector<vector<int> > path(m,vector<int>(n));
        for(int i = 0; i < n; ++i) {
            if(g[0][i] == 1) {
                for(;i<n;++i) {
                    path[0][i] = 0;
                }
                break;
            }
            path[0][i] = 1;
        }
        for (int j = 0;j <m;++j) {
            if(g[j][0] == 1) {
                for(;j<m;++j) {
                    path[j][0] = 0;
                }
                break;
            }
            path[j][0] = 1;
        }
        for(int i =1;i<m;++i) {
            for(int j =1;j<n;++j) {
                if(g[i][j] == 1) 
                    path[i][j] = 0;
                else {
                    path[i][j] = path[i-1][j] + path[i][j-1];
                }
            }
        }
        return path[m-1][n-1];
        
    }
};

　　话说一遍ac的感觉真好啊~