leetcode 62. Unique Paths

/*leetcode 62. Unique PathsA 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.解题思路：动态规划用dp[i][j]表示到(i,j)位置的步数，那么:dp[i][j] = dp[i - 1][j] + dp[i][j - 1]注意边界：当i==0||j==0时，dp[i][j] = 1*/#include<iostream>#include <vector>using namespace std;class Solution {public:    int uniquePaths(int m, int n)     {        vector<vector<int>> dp;        dp.resize(m, vector<int>(n));        for (int i = 0; i < m; ++i)        {            for (int j = 0; j < n; ++j)                if (i == 0 || j == 0)                    dp[i][j] = 1;                else                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];        }        return dp[m - 1][n - 1];    }    //优化：节省空间    int uniquePaths1(int m, int n)    {        vector<int> dp(n, 1);        for (int i = 1; i < m; ++i)            for (int j = 1; j < n; ++j)                dp[j] = dp[j] + dp[j - 1];//第i行第j列的步数之和        return dp[n-1];    }};void test(){    Solution sol;    cout << sol.uniquePaths(2, 3) << endl;    cout << sol.uniquePaths1(2, 3) << endl;}int main(){    test();    return 0;}