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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?
分析：这道题可以用动态规划解，f[i][j]表示从(0,0)到(i,j)的路径数，递推公式为f[i][j] = f[i-1][j] + f[i][j-1]。时间复杂度O(n^2)，空间复杂度O(n^2)。代码如下：
class Solution { public: int uniquePaths(int m, int n) { int f[m][n]; for(int i = 0; i < n; i++) f[0][i] = 1; for(int i = 0; i < m; i++) f[i][0] = 1; for(int i = 1; i < m; i++) for(int j = 1; j < n; j++) f[i][j] = f[i][j-1] + f[i-1][j]; return f[m-1][n-1]; } };
其实在用动态规划时，我们可以用一个滚动数组代替二维数组。代码如下：
class Solution { public: int uniquePaths(int m, int n) { vector<int> f(n,0); f[0] = 1; for(int i = 0; i < m; i++) for(int j = 1; j < n; j++) f[j] = f[j] + f[j-1]; return f[n-1]; } };
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原文地址：http://www.cnblogs.com/Kai-Xing/p/4222638.html
