[Leetcode] 61. Unique Paths

题目

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.

解题之法

class Solution {
public:
    int uniquePaths(int m, int n) {

        vector<vector<int> > dp;
   
        for(int k = 0; k< m; k++){     
            vector<int> cc(n, 1);
            dp.push_back(cc);            
        }

        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];
    }
};

分析

这道题让求所有不同的路径的个数，这跟之前那道 Climbing Stairs 爬梯子问题 很类似，那道题是说可以每次能爬一格或两格，问到达顶部的所有不同爬法的个数。而这道题是每次可以向下走或者向右走，求到达最右下角的所有不同走法的个数。那么跟爬梯子问题一样，我们需要用动态规划Dynamic Programming来解，我们可以维护一个二维数组dp，其中dp[i][j]表示到当前位置不同的走法的个数，然后可以得到递推式为: dp[i][j] = dp[i - 1][j] + dp[i][j - 1]。