[leetcode] 780. Reaching Points @ python

原题

A move consists of taking a point (x, y) and transforming it to either (x, x+y) or (x+y, y).

Given a starting point (sx, sy) and a target point (tx, ty), return True if and only if a sequence of moves exists to transform the point (sx, sy) to (tx, ty). Otherwise, return False.

Examples:
Input: sx = 1, sy = 1, tx = 3, ty = 5
Output: True
Explanation:
One series of moves that transforms the starting point to the target is:
(1, 1) -> (1, 2)
(1, 2) -> (3, 2)
(3, 2) -> (3, 5)

Input: sx = 1, sy = 1, tx = 2, ty = 2
Output: False

Input: sx = 1, sy = 1, tx = 1, ty = 1
Output: True

Note:

sx, sy, tx, ty will all be integers in the range [1, 10^9].

解法

递归.
先找出base case: 由于sx, sy都是正数, 转化之后的tx, ty不可能小于sx, sy. 当tx < sx或ty < sy时, 返回False.
(x, y) 可以转化为 (x, kx + y), 也可以转化为(ky+x, y), 所以当tx == sx时, 如果y的变化值是x的倍数, 返回True. 反之亦然.
当x和y两者都变大了, 我们从tx, ty反向推导sx, sy. 当tx > ty时, tx是ty经过若干转化而得, 因此求tx除以ty的余数, 当tx < ty时, ty是tx经过若干转化而得, 因此求ty除以tx的余数. 如此进行递归.
Time: O(n)
Space: O(1)

代码

class Solution(object):
    def reachingPoints(self, sx, sy, tx, ty):
        """
        :type sx: int
        :type sy: int
        :type tx: int
        :type ty: int
        :rtype: bool
        """
        # base case
        if tx < sx or ty < sy:
            return False
        if tx == sx and (ty - sy)%sx == 0:
            return True
        if ty == sy and (tx - sx)%sy == 0:
            return True
        # recursion
        if tx < ty: 
            return self.reachingPoints(sx, sy, tx, ty%tx)
        else:
            return self.reachingPoints(sx, sy, tx%ty, ty)