leetcode [62] 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 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

我一开始采用回溯法做，但是超时了。

C++超时回溯法：

class Solution {
public:
    void dfs(int m,int n,int x,int y,int&count){
        if(x>=m||x<0||y<0||y>=n) return;
        if(x==m-1&&y==n-1) count++;
        dfs(m,n,x+1,y,count);
        dfs(m,n,x,y+1,count);
    }
    int uniquePaths(int m, int n) {
        int x=0,y=0,count=0;
        dfs(m,n,x,y,count);
        return count;
    }
};

后来一想这不就是特别简单的动态规划嘛，dp[i][j]=dp[i-1][j]+dp[i][j-1],(i,j)位置上的路线总数取决于(i,j-1)和(i-1,j)位置上的路线总和。

class Solution {
private:
    int dp[105][105];
public:
    int uniquePaths(int m, int n) {
        for(int i=0;i<m;i++) dp[i][0]=1;
        for(int j=0;j<n;j++) dp[0][j]=1;
        for(int i=1;i<m;i++){
            for(int j=1;j<n;j++){
                dp[i][j]=dp[i-1][j]+dp[i][j-1];
            }
        }
        return dp[m-1][n-1];
    }
};