[LeetCode]--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.

分析

        根据题意，机器人每次只能向右走或向下走，那么机器人每走一步所经的路径数可以由其上一步的路径数决定，而上一步有两种可能，一种可能是上一步为其左边的格子，另一种可能是上一步为其上边的格子，这很显然是一个动态规划问题。设机器人到达位置(i, j)的路径数为f(i, j)，则f(i, j)=f(i-1, j) + f(i, j-1)，注意这里要将每一个格子的路径数初始化为1。算法复杂度为O(m*n)。

解答

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