[LeetCode] 62. Unique Paths

题：https://leetcode.com/problems/unique-paths/description/
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?


思路
动态规划问题，规划模型：
Since the robot can only move right and down, when it arrives at a point, there are only two possibilities:
It arrives at that point from above (moving down to that point);
It arrives at that point from left (moving right to that point).
Thus, we have the following state equations: suppose the number of paths to arrive at a point 

(i, j)

 is denoted as 

P[i][j]

, it is easily concluded that 

P[i][j] = P[i - 1][j] + P[i][j - 1]

.
initialize 

P[0][j] = 1, P[i][0] = 1

 for all valid 

i, j

. Note the initial value is 

1

这里坐标从(0,0)开始
Codeclass Solution {
public:
int uniquePaths(int m, int n) {
//vector<vector<int> > path(m, vector<int> (n, 1));
int path[m]
;
for(int j = 0;j<m;j++) path[j][0] = 1;
for(int j = 0;j<n;j++) path[0][j] = 1;
for (int i = 1; i < m; i++)
for (int j = 1; j < n; j++)
path[i][j] = path[i - 1][j] + path[i][j - 1];
return path[m - 1][n - 1];
}
};开始 用递归的方法做，出现了 超时的问题。下次该类问题 转为 状态转移的，应该多考虑用DP方式解决。
Codeclass Solution {
public:
int pathnums(int m,int n){
if(m==1||n==1){
return 1;
}
int sum = 0;
for(int i = 1; i<=m-1;i++){
sum+=pathnums(i,n-1);

}
for(int j = 1; j<=n-1 ;j++){
sum+=pathnums(m-1,j);

}
return sum;
}

int uniquePaths(int m, int
aa4c
n) {
return pathnums(m,n);
}
};