题目:62. Unique Paths
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?
分析:
可以类比那道Climbing Stairs 爬梯子问题,而这道题是每次可以向下走或者向右走,求到达最右下角的所有不同走法的个数。那么跟爬梯子问题一样,我们需要用动态规划Dynamic Programming来解。
我们可以维护一个二维数组dp,其中dp[i][j]表示到当前位置不同的走法的个数,然后可以得到递推式为:
dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
代码:
class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int> > dp(m, vector<int>(n, 1));
for(int i = 1; i < m; i ++)
{
for(int j = 1; j < n; j ++)
{
dp[i][j] = dp[i-1][j] + dp[i][j-1];
}
}
return dp[m-1][n-1];
}
};