Ifzliters of water is measurable, you must havezliters of water contained withinone or both bucketsby 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...
学习leetcode_365: Water and Jug Problem的解法:辗转相除相关内容(最大公约数、裴蜀定理、欧几里得算法和扩展欧几里得算法)。 正文: 1、问题描述 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 possibl...
Python Code to Solve Water Jug Problem fromcollectionsimportdequedefwater_jug_problem_trace_path(jug1_capacity,jug2_capacity,target):# Queue to keep track of states and the path to reach themqueue=deque([((0,0),[(0,0)])])# Set to keep track of visited statesvisited=set()visited.add...
学习leetcode_365: Water and Jug Problem的解法:辗转相除相关内容(最大公约数、裴蜀定理、欧几里得算法和扩展欧几里得算法)。 正文: 1、问题描述 You are given two jugs with capacitiesxandylitres. There is an infinite amount of water supply available. You need to determine whether it is possible to m...
摘要:学习leetcode_365: Water and Jug Problem的解法:辗转相除相关内容(最大公约数、裴蜀定理、欧几里得算法和扩展欧几里得算法)。正文:1、问题描述You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. ...
defcanMeasureWater(x,y,z):#首先处理几种极端情况ifx+y<z:returnFalse elif z==x or z==y:returnTrue elif x==0or y==0:returnFalse #判断z是否能整除最大公约数returnz%math.gcd(x,y)==0#math.gcd()这个函数就是直接求出参数的最大公约数 ...
class Solution(object): def trap(self, height): stack = [] water = 0 i=0 while i<len(height): if len(stack) == 0 or height[stack[-1]]>=height[i]: stack.append(i) i+=1 else: x = stack[-1] stack.pop() if len(stack) != 0: temp = min(height[stack[-1]],height[i...
0365.Water and Jug Problem.py 0366.Find Leaves of Binary Tree.py 0367.Valid Perfect Square.py 0368.Largest Divisible Subset.py 0369.Plus One Linked List.py 0370.Range Addition.py 0371.Sum of Two Integers.py 0372.Super Pow.py 0373.Find K Pairs with Smallest Sums.py 0374.Guess Number Hi...
365 Water and Jug Problem C++ Python O(logn) O(1) Medium Euclidean Algorithm 372 Super Pow C++ Python O(n) O(1) Medium 382 Linked List Random Node C++ Python O(n) O(1) Medium Reservoir Sampling 386 Lexicographical Numbers C++ Python O(n) O(1) Medium 390 Elimination Game C++ Pyth...
water -- 166 times other -- 165 times without -- 161 times about -- 161 times how -- 160 times these -- 159 times much -- 151 times woods -- 151 times through -- 151 times yet -- 150 times see -- 148 times long -- 146 times made -- 141 times did -- 137 times those -...