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feat: add solutions to lc problem: No.0365 (doocs#2266)
No.0365.Water and Jug Problem
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‎solution/0300-0399/0365.Water and Jug Problem/README.md

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@@ -51,187 +51,134 @@
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## 解法
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### 方法一
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### 方法一：DFS
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我们不妨记 $jug1Capacity$ 为 $x$, $jug2Capacity$ 为 $y$, $targetCapacity$ 为 $z$。
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接下来，我们设计一个函数 $dfs(i, j)$，表示当前 $jug1$ 中有 $i$ 升水，$jug2$ 中有 $j$ 升水，是否可以得到 $z$ 升水。
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函数 $dfs(i, j)$ 的执行过程如下：
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- 如果 $(i, j)$ 已经被访问过，返回 $false$。
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- 如果 $i = z$ 或者 $j = z$ 或者 $i + j = z$，返回 $true$。
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- 如果我们给 $jug1$ 倒满水，或者给 $jug2$ 倒满水，或者将 $jug1$ 清空，或者将 $jug2$ 清空，可以得到 $z$ 升水，返回 $true$。
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- 如果我们将 $jug1$ 中的水倒入 $jug2$，或者将 $jug2$ 中的水倒入 $jug1$，可以得到 $z$ 升水，返回 $true$。
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答案即为 $dfs(0, 0)$。
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时间复杂度 $O(x + y)$，空间复杂度 $O(x + y)$。其中 $x$ 和 $y$ 分别为 $jug1Capacity$ 和 $jug2Capacity$ 的大小。
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<!-- tabs:start -->
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```python
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class Solution:
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def canMeasureWater(
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self, jug1Capacity: int, jug2Capacity: int, targetCapacity: int
62-
) -> bool:
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stk, seen = [], set()
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stk.append([0, 0])
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66-
def get_hash(nums):
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return nums[0] * 10000006 + nums[1]
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69-
while stk:
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if get_hash(stk[-1]) in seen:
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stk.pop()
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continue
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seen.add(get_hash(stk[-1]))
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cur = stk.pop()
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cur1, cur2 = cur[0], cur[1]
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if (
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cur1 == targetCapacity
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or cur2 == targetCapacity
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or cur1 + cur2 == targetCapacity
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):
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def canMeasureWater(self, x: int, y: int, z: int) -> bool:
76+
def dfs(i: int, j: int) -> bool:
77+
if (i, j) in vis:
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return False
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vis.add((i, j))
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if i == z or j == z or i + j == z:
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return True
82-
stk.append([jug1Capacity, cur2])
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stk.append([0, cur2])
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stk.append([cur1, jug2Capacity])
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stk.append([cur1, 0])
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if cur1 + cur2 > jug1Capacity:
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stk.append([jug1Capacity, cur2 - jug1Capacity + cur1])
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else:
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stk.append([cur1 + cur2, 0])
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if cur1 + cur2 > jug2Capacity:
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stk.append([cur1 - jug2Capacity + cur2, jug2Capacity])
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else:
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stk.append([0, cur1 + cur2])
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return False
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if dfs(x, j) or dfs(i, y) or dfs(0, j) or dfs(i, 0):
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return True
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a = min(i, y - j)
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b = min(j, x - i)
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return dfs(i - a, j + a) or dfs(i + b, j - b)
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vis = set()
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return dfs(0, 0)
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```
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```java
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class Solution {
94+
private Set<Long> vis = new HashSet<>();
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private int x, y, z;
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public boolean canMeasureWater(int jug1Capacity, int jug2Capacity, int targetCapacity) {
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Deque<int[]> stk = new ArrayDeque<>();
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stk.add(new int[] {0, 0});
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Set<Long> seen = new HashSet<>();
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while (!stk.isEmpty()) {
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if (seen.contains(hash(stk.peek()))) {
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stk.pop();
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continue;
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}
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int[] cur = stk.pop();
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seen.add(hash(cur));
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int cur1 = cur[0], cur2 = cur[1];
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if (cur1 == targetCapacity || cur2 == targetCapacity || cur1 + cur2 == targetCapacity) {
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return true;
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}
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stk.offer(new int[] {jug1Capacity, cur2});
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stk.offer(new int[] {0, cur2});
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stk.offer(new int[] {cur1, jug1Capacity});
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stk.offer(new int[] {cur2, 0});
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if (cur1 + cur2 > jug1Capacity) {
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stk.offer(new int[] {jug1Capacity, cur2 - jug1Capacity + cur1});
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} else {
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stk.offer(new int[] {cur1 + cur2, 0});
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}
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if (cur1 + cur2 > jug2Capacity) {
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stk.offer(new int[] {cur1 - jug2Capacity + cur2, jug2Capacity});
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} else {
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stk.offer(new int[] {0, cur1 + cur2});
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}
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x = jug1Capacity;
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y = jug2Capacity;
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z = targetCapacity;
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return dfs(0, 0);
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}
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private boolean dfs(int i, int j) {
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long st = f(i, j);
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if (!vis.add(st)) {
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return false;
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}
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return false;
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if (i == z || j == z || i + j == z) {
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return true;
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}
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if (dfs(x, j) || dfs(i, y) || dfs(0, j) || dfs(i, 0)) {
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return true;
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}
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int a = Math.min(i, y - j);
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int b = Math.min(j, x - i);
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return dfs(i - a, j + a) || dfs(i + b, j - b);
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}
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public long hash(int[] nums) {
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return nums[0] * 10000006L + nums[1];
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private long f(int i, int j) {
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return i * 1000000L + j;
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}
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}
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```
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```cpp
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class Solution {
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public:
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bool canMeasureWater(int jug1Capacity, int jug2Capacity, int targetCapacity) {
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if (jug1Capacity + jug2Capacity < targetCapacity) return false;
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if (jug1Capacity == 0 || jug2Capacity == 0)
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return targetCapacity == 0 || jug1Capacity + jug2Capacity == targetCapacity;
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return targetCapacity % gcd(jug1Capacity, jug2Capacity) == 0;
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}
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148-
int gcd(int a, int b) {
149-
return b == 0 ? a : gcd(b, a % b);
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bool canMeasureWater(int x, int y, int z) {
130+
using pii = pair<int, int>;
131+
stack<pii> stk;
132+
stk.emplace(0, 0);
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auto hash_function = [](const pii& o) { return hash<int>()(o.first) ^ hash<int>()(o.second); };
134+
unordered_set<pii, decltype(hash_function)> vis(0, hash_function);
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while (stk.size()) {
136+
auto st = stk.top();
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stk.pop();
138+
if (vis.count(st)) {
139+
continue;
140+
}
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vis.emplace(st);
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auto [i, j] = st;
143+
if (i == z || j == z || i + j == z) {
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return true;
145+
}
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stk.emplace(x, j);
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stk.emplace(i, y);
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stk.emplace(0, j);
149+
stk.emplace(i, 0);
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int a = min(i, y - j);
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int b = min(j, x - i);
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stk.emplace(i - a, j + a);
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stk.emplace(i + b, j - b);
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}
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return false;
150156
}
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};
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```
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```go
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func canMeasureWater(jug1Capacity int, jug2Capacity int, targetCapacity int) bool {
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if jug1Capacity+jug2Capacity < targetCapacity {
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return false
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}
159-
if jug1Capacity == 0 || jug2Capacity == 0 {
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return targetCapacity == 0 || jug1Capacity+jug2Capacity == targetCapacity
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}
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var gcd func(a, b int) int
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gcd = func(a, b int) int {
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if b == 0 {
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return a
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func canMeasureWater(x int, y int, z int) bool {
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type pair struct{ x, y int }
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vis := map[pair]bool{}
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var dfs func(int, int) bool
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dfs = func(i, j int) bool {
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st := pair{i, j}
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if vis[st] {
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return false
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}
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return gcd(b, a%b)
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vis[st] = true
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if i == z || j == z || i+j == z {
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return true
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}
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if dfs(x, j) || dfs(i, y) || dfs(0, j) || dfs(i, 0) {
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return true
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}
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a := min(i, y-j)
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b := min(j, x-i)
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return dfs(i-a, j+a) || dfs(i+b, j-b)
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}
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return targetCapacity%gcd(jug1Capacity, jug2Capacity) == 0
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}
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```
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```cs
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using System;
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public class Solution {
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public bool CanMeasureWater(int x, int y, int z) {
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if (x == 0 || y == 0) return z == x || z == y;
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var gcd = GetGcd(x, y);
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return z >= 0 && z <= x + y && z % gcd == 0;
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}
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private int GetGcd(int x, int y)
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{
186-
while (x > 0)
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{
188-
var quotient = x / y;
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var reminder = x % y;
190-
if (reminder == 0)
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{
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return y;
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}
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x = y;
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y = reminder;
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}
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throw new Exception("Invalid x or y");
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}
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}
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```
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<!-- tabs:end -->
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### 方法二
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<!-- tabs:start -->
207-
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```python
209-
class Solution:
210-
def canMeasureWater(
211-
self, jug1Capacity: int, jug2Capacity: int, targetCapacity: int
212-
) -> bool:
213-
if jug1Capacity + jug2Capacity < targetCapacity:
214-
return False
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if jug1Capacity == 0 or jug2Capacity == 0:
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return targetCapacity == 0 or jug1Capacity + jug2Capacity == targetCapacity
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return targetCapacity % gcd(jug1Capacity, jug2Capacity) == 0
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```
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220-
```java
221-
class Solution {
222-
public boolean canMeasureWater(int jug1Capacity, int jug2Capacity, int targetCapacity) {
223-
if (jug1Capacity + jug2Capacity < targetCapacity) {
224-
return false;
225-
}
226-
if (jug1Capacity == 0 || jug2Capacity == 0) {
227-
return targetCapacity == 0 || jug1Capacity + jug2Capacity == targetCapacity;
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}
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return targetCapacity % gcd(jug1Capacity, jug2Capacity) == 0;
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}
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232-
private int gcd(int a, int b) {
233-
return b == 0 ? a : gcd(b, a % b);
234-
}
181+
return dfs(0, 0)
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}
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```
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