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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.
代码：
class Solution { public: int uniquePaths(int m, int n) { int dp[m][n]; memset(dp, 0, sizeof(dp)); for ( size_t i = 0; i < n; ++i ) dp[0][i] = 1; for ( size_t i = 0; i < m; ++i ) dp[i][0] = 1; for ( size_t i = 1; i < m; ++i ) { for ( size_t j = 1; j < n; ++j ) { dp[i][j] = dp[i-1][j] + dp[i][j-1]; } } return dp[m-1][n-1]; } };
tips：
常规dp解法。
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上面的代码有可以改进的地方：dp[m][n]并不用这些额外空间，只需要两个长度为n的数组即可；一个保存前一行的状态，一个用于遍历当前行的状态，每次滚动更新，可以省去额外空间。沿着上述思路改进了一版代码如下：
class Solution { public: int uniquePaths(int m, int n) { int curr[n], pre[n]; for ( size_t i = 0; i<n; ++i ) { pre[i]=1; curr[i]=0; } curr[0] = 1; for ( size_t i = 1; i<m; ++i ) { for ( size_t j = 1; j<n; ++j ) { curr[j] = curr[j-1] + pre[j]; pre[j] = curr[j]; } curr[0] = 1; } return pre[n-1]; } };
这个代码空间复杂度降到了O(n)，但还是可以改进。其实只用一个一维的数组dp就可以了，代码如下。
class Solution { public: int uniquePaths(int m, int n) { int curr[n]; memset(curr, 0, sizeof(curr)); curr[0] = 1; for ( size_t i = 0; i < m; ++i ) { for ( size_t j = 1; j < n; ++j ) { curr[j] = curr[j-1] + curr[j]; } } return curr[n-1]; } };
这里用到了滚动数组的技巧。有个细节需要注意，外层dp是可以从0行开始，省去了一部分代码。
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4530940.html
