【Leetcode】【python】Unique Paths/Unique Paths II
2017-10-14 04:18
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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),然后循环计算出所有其他的。
代码
class Solution(object): def uniquePaths(self, m, n): """ :type m: int :type n: int :rtype: int """ dp = [[1 for __ in range(n)] for __ in range(m)] # print dp for i in range(1, n): for j in range(1, m): dp[j][i] = dp[j - 1][i] + dp[j][i - 1] return dp[m - 1][n - 1]
Unique Paths II
题目大意
如果道路上有障碍,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。0表示道路通行,1表示有障碍。解题思路
和上题区别仅仅是有障碍,那么有障碍的地方就是0(不可能走到),然后初始化一下第一行和第一列(若其中有障碍则之后皆为0),最后还是上题那样做。代码
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 m = len(obstacleGrid) # 纵向 n = len(obstacleGrid[0]) # 横向 dp = [[0 for __ in range(n)] for __ in range(m)] # 初始化,所有为0 # 第一个可以走设置为1 dp[0][0] = 1 # 检查第一行和第一列是否有障碍 for i in range(1, m): dp[i][0] = dp[i - 1][0] if obstacleGrid[i][0] == 0 else 0 for j in range(1, n): dp[0][j] = dp[0][j - 1] if obstacleGrid[0][j] == 0 else 0 for i in range(1, m): for j in range(1, n): if obstacleGrid[i][j] == 1: dp[i][j] = 0 else: dp[i][j] = dp[i - 1][j] + dp[i][j - 1] return dp[m - 1][n - 1]
总结
该题动态规划状态转移方程式:dp[i][j] = dp[i - 1][j] + dp[i][j - 1]
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