209 lines
5.7 KiB
Markdown
209 lines
5.7 KiB
Markdown
|
|
|||
|
|
## 题目地址
|
|||
|
|
https://leetcode.com/problems/water-and-jug-problem/description/
|
|||
|
|
|
|||
|
|
## 题目描述
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
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 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
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
|
|||
|
|
## BFS（超时）
|
|||
|
|
|
|||
|
|
### 思路
|
|||
|
|
|
|||
|
|
两个水壶的水我们考虑成状态，然后我们不断进行倒的操作，改变状态。那么初始状态就是（0 0） 目标状态就是 （any, z）或者 (z, any)，其中any 指的是任意升水。
|
|||
|
|
|
|||
|
|
|
|||
|
|
已题目的例子，其过程示意图，其中括号表示其是由哪个状态转移过来的：
|
|||
|
|
|
|||
|
|
0 0
|
|||
|
|
3 5(0 0) 3 0 (0 0 )0 5(0 0)
|
|||
|
|
3 2(0 5) 0 3(0 0)
|
|||
|
|
0 2(3 2)
|
|||
|
|
2 0(0 2)
|
|||
|
|
2 5(2 0)
|
|||
|
|
3 4(2 5) bingo
|
|||
|
|
|
|||
|
|
### 代码
|
|||
|
|
|
|||
|
|
```python
|
|||
|
|
class Solution:
|
|||
|
|
def canMeasureWater(self, x: int, y: int, z: int) -> bool:
|
|||
|
|
if x + y < z:
|
|||
|
|
return False
|
|||
|
|
queue = [(0, 0)]
|
|||
|
|
seen = set((0, 0))
|
|||
|
|
|
|||
|
|
while(len(queue) > 0):
|
|||
|
|
a, b = queue.pop(0)
|
|||
|
|
if a ==z or b == z or a + b == z:
|
|||
|
|
return True
|
|||
|
|
states = set()
|
|||
|
|
|
|||
|
|
states.add((x, b))
|
|||
|
|
states.add((a, y))
|
|||
|
|
states.add((0, b))
|
|||
|
|
states.add((a, 0))
|
|||
|
|
states.add((min(x, b + a), 0 if b < x - a else b - (x - a)))
|
|||
|
|
states.add((0 if a + b < y else a - (y - b), min(b + a, y)))
|
|||
|
|
for state in states:
|
|||
|
|
if state in seen:
|
|||
|
|
continue;
|
|||
|
|
queue.append(state)
|
|||
|
|
seen.add(state)
|
|||
|
|
return False
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
**复杂度分析**
|
|||
|
|
|
|||
|
|
- 时间复杂度：由于状态最多有$O((x + 1) * (y + 1))$ 种，因此总的时间复杂度为$O(x * y)$。
|
|||
|
|
- 空间复杂度：我们使用了队列来存储状态，set 存储已经访问的元素，空间复杂度和状态数目一致，因此空间复杂度是$O(x * y)$。
|
|||
|
|
|
|||
|
|
上面的思路很直观，但是很遗憾这个算法在 LeetCode 的表现是 TLE(Time Limit Exceeded)。不过如果你能在真实面试中写出这样的算法，我相信大多数情况是可以过关的。
|
|||
|
|
|
|||
|
|
我们来看一下有没有别的解法。实际上，上面的算法就是一个标准的 BFS。如果从更深层次去看这道题，会发现这道题其实是一道纯数学问题，类似的纯数学问题在 LeetCode 中也会有一些，不过大多数这种题目，我们仍然可以采取其他方式 AC。那么让我们来看一下如何用数学的方式来解这个题。
|
|||
|
|
|
|||
|
|
## 数学法 - 最大公约数
|
|||
|
|
|
|||
|
|
### 思路
|
|||
|
|
|
|||
|
|
这是一道关于`数论`的题目，确切地说是关于`裴蜀定理`（英语：Bézout's identity）的题目。
|
|||
|
|
|
|||
|
|
摘自wiki的定义：
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
对任意两个整数 a、b，设 d是它们的最大公约数。那么关于未知数 x和 y的线性丢番图方程（称为裴蜀等式）：
|
|||
|
|
|
|||
|
|
ax+by=m
|
|||
|
|
|
|||
|
|
有整数解 (x,y) 当且仅当 m是 d的整数倍。裴蜀等式有解时必然有无穷多个解。
|
|||
|
|
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
因此这道题可以完全转化为`裴蜀定理`。还是以题目给的例子`x = 3, y = 5, z = 4`，我们其实可以表示成`3 * 3 - 1 * 5 = 4`, 即`3 * x - 1 * y = z`。我们用a和b分别表示3
|
|||
|
|
升的水壶和5升的水壶。那么我们可以：
|
|||
|
|
|
|||
|
|
|
|||
|
|
- 倒满a（**1**）
|
|||
|
|
- 将a倒到b
|
|||
|
|
- 再次倒满a（**2**）
|
|||
|
|
- 再次将a倒到b（a这个时候还剩下1升）
|
|||
|
|
- 倒空b（**-1**）
|
|||
|
|
- 将剩下的1升倒到b
|
|||
|
|
- 将a倒满（**3**）
|
|||
|
|
- 将a倒到b
|
|||
|
|
- b此时正好是4升
|
|||
|
|
|
|||
|
|
上面的过程就是`3 * x - 1 * y = z`的具体过程解释。
|
|||
|
|
|
|||
|
|
**也就是说我们只需要求出x和y的最大公约数d，并判断z是否是d的整数倍即可。**
|
|||
|
|
|
|||
|
|
|
|||
|
|
### 代码
|
|||
|
|
|
|||
|
|
代码支持：Python3，JavaScript
|
|||
|
|
|
|||
|
|
|
|||
|
|
Python Code：
|
|||
|
|
|
|||
|
|
```python
|
|||
|
|
class Solution:
|
|||
|
|
def canMeasureWater(self, x: int, y: int, z: int) -> bool:
|
|||
|
|
if x + y < z:
|
|||
|
|
return False
|
|||
|
|
|
|||
|
|
if (z == 0):
|
|||
|
|
return True
|
|||
|
|
|
|||
|
|
if (x == 0):
|
|||
|
|
return y == z
|
|||
|
|
|
|||
|
|
if (y == 0):
|
|||
|
|
return x == z
|
|||
|
|
|
|||
|
|
def GCD(a, b):
|
|||
|
|
smaller = min(a, b)
|
|||
|
|
while smaller:
|
|||
|
|
if a % smaller == 0 and b % smaller == 0:
|
|||
|
|
return smaller
|
|||
|
|
smaller -= 1
|
|||
|
|
|
|||
|
|
return z % GCD(x, y) == 0
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
JavaScript：
|
|||
|
|
|
|||
|
|
|
|||
|
|
```js
|
|||
|
|
/**
|
|||
|
|
* @param {number} x
|
|||
|
|
* @param {number} y
|
|||
|
|
* @param {number} z
|
|||
|
|
* @return {boolean}
|
|||
|
|
*/
|
|||
|
|
var canMeasureWater = function(x, y, z) {
|
|||
|
|
if (x + y < z) return false;
|
|||
|
|
|
|||
|
|
if (z === 0) return true;
|
|||
|
|
|
|||
|
|
if (x === 0) return y === z;
|
|||
|
|
|
|||
|
|
if (y === 0) return x === z;
|
|||
|
|
|
|||
|
|
function GCD(a, b) {
|
|||
|
|
let min = Math.min(a, b);
|
|||
|
|
while (min) {
|
|||
|
|
if (a % min === 0 && b % min === 0) return min;
|
|||
|
|
min--;
|
|||
|
|
}
|
|||
|
|
return 1;
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
return z % GCD(x, y) === 0;
|
|||
|
|
};
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
实际上求最大公约数还有更好的方式，比如辗转相除法：
|
|||
|
|
|
|||
|
|
```python
|
|||
|
|
def GCD(a, b):
|
|||
|
|
if b == 0: return a
|
|||
|
|
return GCD(b, a % b)
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
**复杂度分析**
|
|||
|
|
|
|||
|
|
- 时间复杂度：$O(log(max(a, b)))$
|
|||
|
|
- 空间复杂度：空间复杂度取决于递归的深度，因此空间复杂度为 $O(log(max(a, b)))$
|
|||
|
|
|
|||
|
|
|
|||
|
|
## 关键点分析
|
|||
|
|
|
|||
|
|
- 数论
|
|||
|
|
- 裴蜀定理
|
|||
|
|
|
|||
|
|
更多题解可以访问我的LeetCode题解仓库：https://github.com/azl397985856/leetcode 。 目前已经接近30K star啦。
|
|||
|
|
|
|||
|
|
大家也可以关注我的公众号《脑洞前端》获取更多更新鲜的LeetCode题解
|
|||
|
|
|
|||
|
|
|