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...
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.p...
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 [暴力解法]: 时间分...
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 Code publicclassSoluti...
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 ...
A simple solution to group the jugs into pairs is to compare a red jug with each blue jug. Once we get the matching pair, group that pair, and repeat for the next red jug. This algorithm can be easily implemented, but it will perform at mostn2comparisons. ...
英文: She poured some water into the earthenware jug.中文: 她向那个陶罐里倒了些水。英文: Cost: Bottled water can cost approximately $1 for a gallon jug, while tap water costs pennies on the dollar.中文: 花费:瓶装水大约每加仑1美元,而自来水仅需几美分。
首先可以肯定的是只有xy俩水壶,大于x+y的水肯定是量不出来的,这个没啥大问题。重点是为什么只要x和y的最大公约数能整除z就可以量出z呢? 这里我只找到了一个裴蜀定理来解释了。 publicclassSolution{privateintgcd(int x,int y){if(x%y==0)returny;returngcd(y,x%y);}publicbooleancanMeasureWater(int x,...
英语解释 water-jug problem water-jug problem是什么意思、water-jug problem怎么读 water-jug problem汉语翻译 【计】 水和罐问题
class Solution: def canMeasureWater(self, x: int, y: int, z: int) -> bool: if z == 0 or (z <= x + y and z % self.__gcd(x, y) == 0): return True return False def __gcd(self, x: int, y: int) -> int: