LeetCode—Unique Paths

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.

题解：题目使用动态规划思想，result[i][j]表示从(0,0)点开始到(i,j)点的不同路径数，易知result[0][0] = 0;从start(0,0)到第一行或第一列的每个位置都是直走，只有一条路（一直往右走或一直往下走），所以result[0][j]=1,1<=j<=n-1;result[j][0]=1,1<=j<=m-1;对于(i,j)点(1<=i<=m-1,1<=j<=n-1)，到达这一点或者是从上方的点(i-1,j)而来，或者是从左边的点(i,j-1)而来,所以result[i][j] = result[i-1][j]+result[i][j-1],有了递归关系式，代码如下：

class Solution {
public:
    int uniquePaths(int m, int n) {
    int result[m][n];
    result[0][0] = 0;
    if(m==1&&n==1)return 1;
    for(int i = 1;i<=n-1;i++)
    result[0][i] = 1;
    for(int i = 1;i<=m-1;i++)
    result[i][0] = 1;
    
    for(int i = 1;i<m;i++)
    for(int j = 1;j<n;j++)
    {
         result[i][j] = result[i-1][j]+result[i][j-1];
    }
    return result[m-1][n-1];
    }
};