水壶问题
作者:互联网
365. 水壶问题
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
分析一下x = 3, y = 5, z = 4的情况,最终需要4, 5-1=4,所以构造出一个1就可以,5-3=2,3-2=1
- x空,装满y
- y倒入x直到x装满,y剩下2
- x倒掉,y倒入x(此时x为2,y空)
- y装满,倒入x直到x满(此时x为3,y等于4)
- 倒出x
# DFS 搜索操作:清空、装满、倒入
def canMeasureWater(x,y,z):
if x+y<z:
return False
if x ==0 or y == 0:
return z == 0 or x+y == z
stack = [(0,0)]#初始化x,y
seen = set()
while stack:
remain_x,remain_y = stack.pop()
if remain_x+remain_y == z:
return True
if (remain_x,remain_y) in seen:
continue
seen.add((remain_x,remain_y))
stack.append((x, remain_y))# 把 X 壶灌满
stack.append((remain_x, y))# 把 Y 壶灌满
stack.append((0, remain_y))# 把 X 壶倒空
stack.append((remain_x, 0))# 把 Y 壶倒空
# 把 X 壶的水灌进 Y 壶,直至灌满或倒空
stack.append((remain_x - min(remain_x, y - remain_y), remain_y + min(remain_x, y - remain_y)))
# 把 Y 壶的水灌进 X 壶,直至灌满或倒空
stack.append((remain_x + min(remain_y, x - remain_x), remain_y - min(remain_y, x - remain_x)))
return False
# BFS 搜索操作:清空、装满、倒入
def canMeasureWater(x,y,z):
if x+y<z:
return False
if x ==0 or y == 0:
return z == 0 or x+y == z
queue = [(0,0)]#初始化x,y
seen = set()
while queue:
remain_x,remain_y = queue.pop(0)
if remain_x+remain_y == z:
return True
if (remain_x,remain_y) in seen:
continue
seen.add((remain_x,remain_y))
queue.append((x, remain_y))# 把 X 壶灌满
queue.append((remain_x, y))# 把 Y 壶灌满
queue.append((0, remain_y))# 把 X 壶倒空
queue.append((remain_x, 0))# 把 Y 壶倒空
# 把 X 壶的水灌进 Y 壶,直至灌满或倒空
queue.append((remain_x - min(remain_x, y - remain_y), remain_y + min(remain_x, y - remain_y)))
# 把 Y 壶的水灌进 X 壶,直至灌满或倒空
queue.append((remain_x + min(remain_y, x - remain_x), remain_y - min(remain_y, x - remain_x)))
return False
#裴蜀定理
def canMeasureWater(x,y,z):
import math
if x+y < z:
return False
if x ==0 or y == 0:
return z == 0 or x+y == z
return z%math.gcd(x,y) == 0
标签:return,水壶,装满,问题,倒入,canMeasureWater,def 来源: https://www.cnblogs.com/gongyanzh/p/12539082.html