LeetCode | 0365. Water and Jug Problem水壶问题【Python】

LeetCode 0365. Water and Jug Problem水壶问题【Medium】【Python】【BFS】【数学】

Problem

You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs.

If z liters of water is measurable, you must have z liters of water contained within one or both buckets by the end.

Operations allowed:

Fill any of the jugs completely with water.
Empty any of the jugs.
Pour water from one jug into another till the other jug is completely full or the first jug itself is empty.

Example 1: (From the famous “Die Hard” example)

Input: x = 3, y = 5, z = 4
Output: True

Example 2:

Input: x = 2, y = 6, z = 5
Output: False

问题

力扣

有两个容量分别为 x升和 y升的水壶以及无限多的水。请判断能否通过使用这两个水壶，从而可以得到恰好 z升的水？

如果可以，最后请用以上水壶中的一或两个来盛放取得的 z升水。

你允许：

装满任意一个水壶
清空任意一个水壶
从一个水壶向另外一个水壶倒水，直到装满或者倒空

示例 1: (From the famous “Die Hard” example)

输入: x = 3, y = 5, z = 4
输出: True

示例 2:

输入: x = 2, y = 6, z = 5
输出: False

思路

解法一

BFS

每次水壶都有三个操作：加满水、清空水、相互倒。

Python3代码

class Solution:
    def canMeasureWater(self, x: int, y: int, z: int) -> bool:
        # solution one: BFS
        from collections import deque
        queue = deque([[0, 0]])
        visited = set([(0, 0)])

        while queue:
            cur_x, cur_y = queue.pop()
            if z in [cur_x, cur_y, cur_x + cur_y]:
                return True
            for item in [
                # x 加满水，y 加满水
                (x, cur_y), (cur_x, y),
                # x 清空水，y 清空水
                (0, cur_y), (cur_x, 0),
                # 把 x 壶的水灌进 y 壶，直至灌满或倒空
                (cur_x + cur_y - y, y) if cur_x + cur_y >= y else (0, cur_x + cur_y),
                # 把 X 壶的水灌进 Y 壶，直至灌满或倒空
                (x, cur_x + cur_y - x) if cur_x + cur_y >= x else (cur_x + cur_y, 0)]:
                if item not in visited:
                    queue.appendleft(item)  # 从队列左边加入元素
                    visited.add(item)
        return False

解法二

裴蜀定理

能否找到整数 a，b 使得方程 ax + by = z 有解。
有整数解时，当且仅当 z 是 a 和 b 的最大公约数 d 的倍数。

Python3代码

class Solution:
    def canMeasureWater(self, x: int, y: int, z: int) -> bool:
        # solution two: 裴蜀定理
        import math
        if x + y < z:
            return False
        if x == z or y == z or x + y == z:
            return True
        return z % math.gcd(x, y) == 0