LeetCode365. Water and Jug Problem

最新推荐文章于 2024-09-13 08:04:16 发布

tigerhuli

最新推荐文章于 2024-09-13 08:04:16 发布

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本文探讨了使用两个容量分别为x和y升的水壶，是否能精确测量出z升水的问题。通过分析，我们发现能产生的最少量的水是两个壶的最大公倍数，其余可测量的水量以此递增。文章提供了具体的示例和解决方案。

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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

这道题在考察大家的观察能力。

举个例子。如果我的两个壶都是偶数，请问能量出奇数的结果吗？

所以，通过观察发现，我们能产生的最少量的水，其实是两个壶的最大公倍数。然后其余能够测量的水就以最大公倍数递增，直到装满两个壶。

注意输入的壶的容量有0的情况。

class Solution {
private:
	int gcd(int a, int b) {
		if (a < b) swap(a, b);
		while (b > 1) {
			if (a % b == 0) return b;
			int tmp = a;
			a = b;
			b = tmp % a;
		}
		return 1;
	}
public:
	bool canMeasureWater(int x, int y, int z) {
        if(x==0 || y == 0) return x==z || y==z;
		int gcd = this->gcd(x, y);
		if (z % gcd == 0 && z <= x + y) return true;
		else return 0;
	}
};