LeetCode-365 Water and Jug Problem
作者:互联网
题目描述
You are given two jugs with capacities xand 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.
题目大意
两个大小分别为x和y升的水壶,可以无限次的给水壶充满水,并且两个水壶之间可以互相倒水,是否可以实现两个水壶中的水相加等于z。
示例
E1
Input: x = 3, y = 5, z = 4 Output: True
E2
Input: x = 2, y = 6, z = 5 Output: False
解题思路
根据LeetCode@lblbxuxu2的思路,用v来表示当前的获得的水的总量,
当v < x时,能做的有意义的操作只能是将y灌满,因此:v += y
当v > x时,能做的有意义的操作只能是将x清空,因此:v -= x
循环判定是否存在满足v = z的情况。
复杂度分析
时间复杂度:O(N)
空间复杂度:O(1)
代码
class Solution { public: bool canMeasureWater(int x, int y, int z) { if(x + y == z) return true; if(x + y < z) return false; // 若x比y小,则将两个数字互换 if(x < y) { int tmp = x; x = y; y = tmp; } int v = 0; while(1) { if(v < x) v += y; else v -= x; if(v == z) return true; if(v == 0) return false; } } };
标签:water,return,jugs,int,复杂度,Water,水壶,Problem,365 来源: https://www.cnblogs.com/heyn1/p/11244169.html