leetcode 62: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.

思路：

这题，数学上的解法就是C{m+n-2}{n-1}。

我们用学过的算法来解。动态规划无疑是非常简单的一种解法。

用DP[i][j]表示从开始到第i+1行j+1列共有的路径条数。

显然这题满足下列最优子结构：

DP[i][j]=DP[i-1][j]+DP[i][j-1]            i,j>0;

DP[i][j]=1                                          i=1或j=1

最后DP[m-1][n-1]即为到finish处的路径条数。

时间复杂度：O(m*n)

实现如下：

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

最后补上数学解法：

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;    }};