problem:
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.
Array Dynamic Programmingthinking:
（1）这道题相当经典，为什么这样说，且听分析
看到题第一反应想到用DFS遍历每一条路径，计个数即可。但是DFS的时间复杂度是路径数total 的线性表达式，提交显示 TLE
（2）抖机灵想到用递归，f(m,n)=f(m-1,n)+f(m,n-1)，当m或n为1时，路径数为1。 但是，递归法的时间复杂度同DFS法一样，提交也是TLE
（3）正确的解法是DP，状态转移方程 同递归法一样：a[m][n] = a[m-1][n]+a[m][n-1],采用自底向上的思路，时间复杂度和空间复杂度都为O(m*n)
code:
DFS法： TLE
class Solution {
public:
int uniquePaths(int m, int n) {
int count=0;
int x=0,y=0;
dfs(x,y,m,n,count);
return count;
}
protected:
void dfs(int x, int y, int m, int n, int &count)
{
if(x==m-1 && y==n-1)
{
count++;
return;
}
if(x<m-1 && y< n-1)
{
dfs(x+1,y,m,n,count);
dfs(x,y+1,m,n,count);
}
else if(x==m-1 && y<n-1)
dfs(x,y+1,m,n,count);
else
dfs(x+1,y,m,n,count);
}
};class Solution {
public:
int uniquePaths(int m, int n) {
return f(m,n);
}
protected:
int f(int m,int n)
{
if(m==1 || n==1)
return 1;
else
return f(m-1,n)+f(m,n-1);
}
};
class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int> > a(m, vector<int>(n));
for(int i = 0; i < n; i++)
a[0][i] = 1;
for(int i = 0; i < m; i++)
a[i][0] = 1;
for(int i = 1; i < m; i++)
for(int j = 1; j < n; j++)
a[i][j] = a[i-1][j] + a[i][j-1];
return a[m-1][n-1];
}
};
原文：http://blog.csdn.net/hustyangju/article/details/44829339