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.
以下方法，用stack可以遍历每条路径，结果正确，但是Time Limit Exceeded！因为只要得到路径的个数所以并不需要遍历每条路径！思考简单的方法.
class Solution { public: int uniquePaths(int m, int n) { int row=0,col=0,res=0; pair<int,int> rowCol; rowCol = make_pair(row,col); stack<pair<int,int>> stackRowCol; stackRowCol.push(rowCol); while(!stackRowCol.empty()){ pair<int,int> temp = stackRowCol.top(); stackRowCol.pop(); row = temp.first; col = temp.second; if(row==m-1 && col==n-1){ res++; continue;} if(row<m-1) { rowCol = make_pair(row+1,col); stackRowCol.push(rowCol); } if(col<n-1) { rowCol = make_pair(row,col+1); stackRowCol.push(rowCol); } }//end while return res; } };
用动态规划的方法（DP），定义一个m*n大小的矩阵，矩阵里的值表示从当前点到右下角的路径数，假如(i,j)点到终点的路径数是C(i,j),则
C(i,j) = C(i+1,j)+C(i,j+1);由此可以用O（m*n）的时间复杂度和O（m*n）的空间复杂度计算出结果，具体编程如下：
class Solution { public: int uniquePaths(int m, int n) { int row=m-2,col=n-2,res=0; vector<int> vec0(n,1); vector<vector<int> > vec(m,vec0); for(int i=row;i>=0;i--){ for(int j=col;j>=0;j--){ vec[i][j]=vec[i][j+1]+vec[i+1][j]; } } return vec[0][0]; } };
[LeetCode] Unique Paths,布布扣,bubuko.com
原文：http://www.cnblogs.com/Xylophone/p/3867440.html