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:...
classSolution {public:intuniquePathsWithObstacles(vector<vector<int>>&obstacleGrid) {intm =obstacleGrid.size();intn = obstacleGrid[0].size(); vector<vector<int> > a(m, vector<int>(n));intpath1,path2,path=0;intj;for(inti =0; i < m; i++) {if(obstacleGrid[i][0] ==1) {for...
代码如下: classSolution {public:intuniquePathsWithObstacles(vector<vector<int> > &obstacleGrid) {//if(obstacleGrid.size() < 1) return 1;if(obstacleGrid[0][0] ==1)return0;if(obstacleGrid.size() ==1&& obstacleGrid[0].size()==1)return1;intm = obstacleGrid.size(), n = obstacleGrid...
输出: 2 解题思路 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): """ :type obstacleGrid: List[Li...
Unique Paths II -- LeetCode 原题链接:http://oj.leetcode.com/problems/unique-paths-ii/ 这道题跟Unique Paths非常类似,只是这道题给机器人加了障碍,不是每次都有两个选择(向右,向下)了。因为有了这个条件,所以Unique Paths中最后一个直接求组合的方法就不适用了,这里最好的解法就是用动态规划了。递推...
2. 代码 AI检测代码解析 class Solution(object): def uniquePathsWithObstacles(self, obstacleGrid): """ :type obstacleGrid: List[List[int]] :rtype: int """ if obstacleGrid[0][0] == 1: # 如果第一个格子有障碍,就无法走 return 0 ...
class Solution: def uniquePaths(self, m: int, n: int) -> int: # dp[j] 表示从 (0, 0) 到 (i - 1, j - 1) 的不同路径数,初始化均为 0 。 # 这里使用了一维数组 + 临时变量的优化,能将空间复杂度从 O(mn) 降到 O(n) 。 dp: List[int] = [0] * (n + 1) # 为了方便后续...
63. Unique Paths II 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 ...
Unique Paths I:: https://leetcode.com/problems/unique-paths/description/ 解题思路: dp[i][j] = dp[i - 1][j] + dp[i][j-1] 代码如下: class Solution { public int uniquePaths(int m, int n) { int[][] res = new int[m][n]; for(int i = 0; i < m; i++) res[i][0...
Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner.The test cases are generated so that the answer will be less than or equal to 2 * 109.Example 1:Input: m = 3, n = 7 Output: 28 ...