一天一道LeetCode系列
（一）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.
（二）解题
主要思想：对于i，j这一点来说，它到终点的路径数dp[i][j] = dp[i+1][j]+ dp[i][j+1]，这就是状态转移方程，然后利用动态规划来求解！
递归版本
class Solution {
public:
int dp[101][101];//用来标记已经计算过的路径
int uniquePaths(int m, int n) {
dp[m-1][n-1] = 1;
int ret = dfsPath(0,0,m-1,n-1);
return ret;
}
int dfsPath(int pm,int pn,int m ,int n)
{
if(pm==m && pn==n) return 1 ;
int down = 0;
int right = 0;
if(pm+1<=m) down = dp[pm+1][pn]==0?dfsPath(pm+1,pn,m,n):dp[pm+1][pn];//往下走的那一格到终点的路径数
if(pn+1<=n) right = dp[pm][pn+1]==0?dfsPath(pm,pn+1,m,n):dp[pm][pn+1];//往右走的那一格到终点的路径数
dp[pm][pn] = down+right;
return dp[pm][pn];
}
};非递归版本
/*
提示：这个版本画个图可能会更好理解
*/
class Solution {
public:
int uniquePaths(int m, int n) {
int dp[101][101];
for(int i = 0 ; i < m ; i++) dp[i][n-1] = 1;//首先初始化dp
for(int i = 0 ; i < n ; i++) dp[m-1][i] = 1;
if(m==1||n==1) return 1;//特殊情况
for(int i = m-2 ; i>=0 ; i--)
for(int j = n-2 ; j>=0 ; j--)
{
dp[i][j] = dp[i+1][j] + dp[i][j+1];//状态转移方程
}
return dp[0][0];
}
};