LeetCode 0365. Water and Jug Problem水壺問題【Medium】【Python】【BFS】【數學】
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
問題
有兩個容量分別爲 x升 和 y升 的水壺以及無限多的水。請判斷能否通過使用這兩個水壺,從而可以得到恰好 z升 的水?
如果可以,最後請用以上水壺中的一或兩個來盛放取得的 z升 水。
你允許:
- 裝滿任意一個水壺
- 清空任意一個水壺
- 從一個水壺向另外一個水壺倒水,直到裝滿或者倒空
示例 1: (From the famous “Die Hard” example)
輸入: x = 3, y = 5, z = 4
輸出: True
示例 2:
輸入: x = 2, y = 6, z = 5
輸出: False
思路
解法一
BFS
每次水壺都有三個操作:加滿水、清空水、相互倒。
Python3代碼
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 加滿水,y 加滿水
(x, cur_y), (cur_x, y),
# x 清空水,y 清空水
(0, cur_y), (cur_x, 0),
# 把 x 壺的水灌進 y 壺,直至灌滿或倒空
(cur_x + cur_y - y, y) if cur_x + cur_y >= y else (0, cur_x + cur_y),
# 把 X 壺的水灌進 Y 壺,直至灌滿或倒空
(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) # 從隊列左邊加入元素
visited.add(item)
return False
解法二
裴蜀定理
能否找到整數 a,b 使得方程 ax + by = z 有解。
有整數解時,當且僅當 z 是 a 和 b 的最大公約數 d 的倍數。
Python3代碼
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