[LeetCode] Unique Paths

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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?

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可以遍曆每條路徑，結果正確，但是Time Limit Exceeded！因為只要得到路徑的個數所以並不需要遍曆每條路徑！思考簡單的方法.

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;    }};

用動態規劃的方法（DP），定義一個m*n大小的矩陣，矩陣裡的值表示從當前點到右下角的路徑數，假如(i,j)點到終點的路徑數是C(i,j),則

C(i,j) = C(i+1,j)+C(i,j+1);由此可以用O（m*n）的時間複雜度和O（m*n）的空間複雜度計算出結果，具體編程如下：

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];    }};