題目鏈接
tag:

• Medium；
• DP；

question：
??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

思路：
??這道題讓求所有不同的路徑，每次可以向下走或者向右走，求到達最右下角的所有不同走法的個數(shù)。那么跟爬梯子問題一樣，我們可以用動態(tài)規(guī)劃Dynamic Programming來解，我們可以維護一個二維數(shù)組dp，其中dp[i][j]表示到當前位置不同的走法的個數(shù)，然后可以得到遞推式為: dp[i][j] = dp[i - 1][j] + dp[i][j - 1]，為了節(jié)省空間，我們使用一維數(shù)組dp，一行一行的刷新也可以，代碼如下：

class Solution {
public:
    int uniquePaths(int m, int n) {
        /*
        // DP with 1 dimensions array  
        if(m<=0 || n<=0)
            return 0;
        vector<int> dp(n,0);
        dp[0] = 1;
        for(int i=0; i<m; ++i) {
            for(int j=1; j<n; ++j) {
                dp[j] += dp[j-1];
            }
        }
        return dp[n-1];
        */
        
        // DP with 2 dimensions array  
        if(m<=0 || n<=0)
            return 0;
        int dp[m][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];
    }
};