标签：oca ogre init bin leetcode ret com hat 节点
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?
Above is a 7 x 3 grid. How many possible unique paths are there?
Example 1:
Input: m = 3, n = 2 Output: 3 Explanation: From the top-left corner, there are a total of 3 ways to reach the bottom-right corner: 1. Right -> Right -> Down 2. Right -> Down -> Right 3. Down -> Right -> Right
Example 2:
Input: m = 7, n = 3 Output: 28
Constraints:
1 <= m, n <= 1002 * 10 ^ 9.求所有不相同的路径，从左上角走到右下角。走法只能是向下或者向右。
题目中给示例是3 X 2的输入，就是两行三列，走法共3种。
容易看出，所有的步骤可以用一个二叉树存储。例如
1. Right -> Right -> Down
2. Right -> Down -> Right
3. Down -> Right -> Right
画成二叉树就是：
所以问题转化为构建二叉树和统计二叉树叶子节点个数的问题。
class Node:
def __init__(self, left=None, right=None):
self.left = left
self.right = right
class Solution(object):
def uniquePaths(self, m, n):
"""
:type m: int
:type n: int
:rtype: int
"""
def recursive(m, n):
root = Node()
if m == 1 and n == 1:
return None
elif m > 1 and n > 1:
root.right = recursive(m-1, n)
root.left = recursive(m, n-1)
elif m > 1:
root.right = recursive(m - 1, n)
elif n > 1:
root.left = recursive(m, n-1)
return root
def count_leaves(root):
count = 0
if root.left is None and root.right is None:
count += 1
if root.left is not None:
count += count_leaves(root.left)
if root.right is not None:
count += count_leaves(root.right)
return count
root = recursive(m, n)
if not root:
return 1
return count_leaves(root)
a = Solution()
b = a.uniquePaths(1, 1)
print(b)
不过很遗憾，这种算法leetcode OJ跑到 37 / 62 个测试样例时就超时了。
求所有不重复路径, Unique Paths， LeetCode题解（四）
标签：oca ogre init bin leetcode ret com hat 节点
原文地址：https://www.cnblogs.com/importsober/p/13209961.html
