Solution1: class Solution { public: int uniquePaths(int m, int n) { vector<int> f(n, 0); f[0] = 1; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { // f[i][j] = f[i-1][j]+f[i][j-1] f[j] = f[j] + f[j-1]; } } return f[...
How many possible unique paths are there? Above is a 3 x 7 grid. How many possible unique paths are there? Note:mandnwill be at most 100. Show Tags SOLUTION 1: 使用DP解决之。很简单。某一个cell有2种可能到达,从上面来,以及从左边来,只需要把每一个cell的可能数计算好,并且把它们相加即可。
How many possible unique paths are there? Above is a 3 x 7 grid. How many possible unique paths are there? Note:mandnwill be at most 100. Show Tags SOLUTION 1: 使用DP解决之。很简单。某一个cell有2种可能到达,从上面来,以及从左边来,只需要把每一个cell的可能数计算好,并且把它们相加即可。
class Solution: # 递归版本实现,当前位置的路径条数 = 当前位置左边的路径条数 + 当前位置上边的路径条数 def __init__(self): # 结果矩阵,用于存储已经遍历过的位置,减少递归重复 self.res = [] def uniquePathsWithObstacles(self, obstacleGrid): """ :type obstacleGrid: List[List[int]] :rtype:...
class Solution(object): def uniquePaths(self, m, n): """ :type m: int :type n: int :rtype: int """ dp = [[1 for i in range(n)] for i in range(m)] for i in range(1, m): for j in range(1, n): dp[i][j] = dp[i-1][j] + dp[i][j-1] return dp[m-1]...
class Solution(object): def uniquePaths(self, m, n): """ :type m: int :type n: int :rtype: int """ total = m + n - 2 v = n - 1 def permutation(m, n): son = 1 for i in range(m, m - n, -1): son *= i ...
思路跟Unique Paths是一样的,只是要分类讨论一下障碍的情况,假设当前格子是障碍,那么到达该格子的路径数目是0。由于无法到达,假设是普通格子,那么由左边和右边的格子相加。 AC源代码 class Solution(object): def uniquePathsWithObstacles(self, obstacleGrid): ...
public class Solution { public int uniquePathsWithObstacles(int[][] obstacleGrid) { int m = obstacleGrid.length, n = obstacleGrid[0].length; int[][] dp = new int[m][n]; for(int i = 0; i < m; i++){ if(obstacleGrid[i][0] == 1) break; ...
class Solution: @lru_cache def uniquePaths(self, m: int, n: int) -> int: if m == 1 or n == 1: return 1 return self.uniquePaths(m - 1, n) + self.uniquePaths(m, n - 1)关键点 排列组合原理记忆化递归基本动态规划问题空间复杂度可以进一步优化到 O(n), 这会是一...
Unique Paths 62. Unique Paths Description 描述:https://leetcode.com/problems/unique-paths/description/ 题意:m列 × n行 的网格,从左上角第一个到右下角最后一个(只能往右和下这两个方向走),有多少条不重复的路径 Solution 1: (Java) 超时 思路1 刚开始想到的是回溯法,但是在 37/62 测试用......