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.
先直接上题目。这道题目是在做leetcode以来,第一次没有百度任何信息,或者是提示。完全是自己写好了递推式,然后用notepad++写出来的代码,
没有使用任何IDE工具。还是小有成就感的。
当然,做的依旧是动态规划的题目。难度类型上升到了medium。主要是得到了递推式:nums[m][n]=nums[m][n-1]+nums[m-1][n]。
class Solution {
public:
int uniquePaths(int m, int n) {
int nums[m][n];//
//得赋初始值
for(int i=0;i<m;i++)
nums[i][0]=1;
for(int i=0;i<n;i++)
nums[0][i]=1;
if((m-1)==0||(n-1)==0)//讨论的是数组的下标
return 1;
else
{
for(int j=1;j<n;j++)
for(int i=1;i<m;i++)
return nums[i][j]=nums[i][j-1]+nums[i-1][j];
}
}
};