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:...
大概意思:和Unique Paths差不多,但是,在网格中会有些障碍,用1来表示; 方法一:和Unique Paths方法一类似,探究到达全部点的路径数(n=上+右),但是这里,将障碍那点的路径数为0; publicclassSolution{publicintuniquePathsWithObstacles(int[][] obstacleGrid){intm=obstacleGrid.length;intn=obstacleGrid[0].length...
1 class Solution { 2 public: 3 int uniquePathsWithObstacles(vector > &obstacleGrid) { 4 // IMPORTANT: Please reset any member data you declared
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[List[int]] :rtype:...
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...
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 ...
[Leetcode][python]Unique Paths/Unique Paths II Unique Paths 题目大意 机器人从起点到终点有多少条不同的路径,只能向右或者向下走。 解题思路 动态规划 由于只能有向下向右,只有从[1][1]开始的格子需要选择走法,第一行和第一列所有都只有一种走法,所有都设置成1,(这里图方便所有都初始化为1),然后循环计算...
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]...
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 ...
在Unique Paths的基础上,只需要当遇到obstacle的时候将dp[i][j]设为0即可 class Solution { public int uniquePathsWithObstacles(int[][] obstacleGrid) { int m = obstacleGrid.length, n = obstacleGrid[0].length; int[][] dp = new int[m][n]; dp[m - 1][n - 1] = 1; for(int i = ...