【LeetCode】Unique Paths

参考链接

http://www.cnblogs.com/remlostime/archive/2012/11/15/2772263.html

题目描述

Unique Paths

 Total Accepted: 9698 Total
Submissions: 31863My Submissions

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.

题目分析

动态规划
开一个f[m]
的数组，f[i][j] = f[i-1][j] + f[i][j-1]，空间时间复杂度O(m*n)。用滚动数组空间复杂度可降为O(n)

代码示例

/*
编译环境CFree 5.0
博客地址：http://blog.csdn.net/Snowwolf_Yang
*/
#include
#include
using namespace std;
#if 0//超时
class Solution {
public:
int uniquePaths(int m, int n) {
int sum = 0;
uniquePathsCore(m,n,0,0,sum);
return sum;
}
void uniquePathsCore(int m, int n, int i, int j, int &sum)
{
if(i == m-1 && j == n-1)
{
sum++;
//printf("(%d,%d) ",i,j);
//printf("end\n");
return;
}
if(j > f(m, vector(n));

for(int i = 0; i < n; i++)
f[0][i] = 1;

for(int i = 0; i < m; i++)
f[i][0] = 1;

for(int i = 1; i < m; i++)
for(int j = 1; j < n; j++)
f[i][j] = f[i-1][j] + f[i][j-1];

return f[m-1][n-1];
}
};
#endif
void test0()
{
Solution so;
if(so.uniquePaths(1,1) != 1)
printf("------------------------failed\n");
else
printf("------------------------passed\n");

if(so.uniquePaths(1,2) != 1)
printf("------------------------failed\n");
else
printf("------------------------passed\n");

if(so.uniquePaths(1,3) != 1)
printf("------------------------failed\n");
else
printf("------------------------passed\n");

if(so.uniquePaths(2,1) != 1)
printf("------------------------failed\n");
else
printf("------------------------passed\n");

if(so.uniquePaths(2,2) != 2)
printf("------------------------failed\n");
else
printf("------------------------passed\n");

int out = so.uniquePaths(23,12) ;
if(out != 193536720)
printf("------------------------failed %d \n",out);
else
printf("------------------------passed\n");
}
int main(int argc, char *argv[])
{
test0();
return 0;
}

推荐学习C++的资料

C++标准函数库
http://download.csdn.net/detail/chinasnowwolf/7108919

在线C++API查询
http://www.cplusplus.com/

vector使用方法

http://www.cplusplus.com/reference/vector/vector/