62. Unique Paths (Graph; DP)
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: 如果使用递归,那么DFS(i,j)=v[i][j]+max{DFS(i,j-1), DFS(i-1,j)},问题是DFS(i,j)可能会被计算两次,一次来自它右边的节点,一次来自它下面的节点。每个节点都被计算两次,时间复杂度就是指数级的了。解决方法是使用动态规划存储节点信息,避免重复计算。
Above is a 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
class Solution { public: int uniquePaths(int m, int n) { int dp[m][n]; dp[0][0] = 1; for(int i = 0; i< n; i++ ) { dp[0][i] = 1; } for(int i = 0; i< m; i++ ) { dp[i][0] = 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]; } };