feat: add solutions to lc problem: No.0765.Couples Holding Hands

Author: yanglbme

solution/0700-0799/0765.Couples Holding Hands/README.md

@@ -35,27 +35,202 @@
 	<li>可以保证<code>row</code> 是序列&nbsp;<code>0...len(row)-1</code>&nbsp;的一个全排列。</li>
 </ol>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+坐错位置的情况与最少需要交换次数：
+
+- 1 对情侣、2 个座位，不需要交换。
+- 2 对情侣、4 个座位，交换 1 次。
+- 3 对情侣、6 个座位。首先交换 1 次使得其中 1 对情侣坐在一起，剩下 2 对情侣、4 个座位。即需要交换 2 次。
+
+以此类推，得到 `f(n)=n-1`。即：n 对情侣相互坐错位置，最少需要交换 `n-1` 次。
+
+把相互坐错位置的情侣放在一组（同个集合），组内有 n 对情侣就需要 `n-1` 次交换。将 n 对情侣分为 K 组：N1,N2...Nk，有 N1+N2+...+Nk=n。需要交换的次数分别为：N1-1、N2-1、...、Nk-1，则总的最少交换次数为 N1-1+N2-1+...+Nk-1=N1+N2+...+Nk-k=n-k。问题转换为：n 对情侣，根据相互坐错位置的条件分组，共有多少个分组。并查集实现。
+
+模板 1——朴素并查集：
+
+```python
+# 初始化，p存储每个点的祖宗节点
+p = [i for i in range(n)]
+# 返回x的祖宗节点
+def find(x):
+    if p[x] != x:
+        # 路径压缩
+        p[x] = find(p[x])
+    return p[x]
+# 合并a和b所在的两个集合
+p[find(a)] = find(b)
+```
+
+模板 2——维护 size 的并查集：
+
+```python
+# 初始化，p存储每个点的祖宗节点，size只有当节点是祖宗节点时才有意义，表示祖宗节点所在集合中，点的数量
+p = [i for i in range(n)]
+size = [1] * n
+# 返回x的祖宗节点
+def find(x):
+    if p[x] != x:
+        # 路径压缩
+        p[x] = find(p[x])
+    return p[x]
+# 合并a和b所在的两个集合
+size[find(b)] += size[find(a)]
+p[find(a)] = find(b)
+```
+
+模板 3——维护到祖宗节点距离的并查集：
+
+```python
+# 初始化，p存储每个点的祖宗节点，d[x]存储x到p[x]的距离
+p = [i for i in range(n)]
+d = [0] * n
+# 返回x的祖宗节点
+def find(x):
+    if p[x] != x:
+        t = find(p[x])
+        d[x] += d[p[x]]
+        p[x] = t
+    return p[x]
+# 合并a和b所在的两个集合
+p[find(a)] = find(b)
+d[find(a)] = dinstance
+```
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def minSwapsCouples(self, row: List[int]) -> int:
+        n = len(row) >> 1
+        p = [i for i in range(n)]
+
+        def find(x):
+            if p[x] != x:
+                p[x] = find(p[x])
+            return p[x]
+
+        for i in range(0, len(row), 2):
+            a, b = row[i] >> 1, row[i + 1] >> 1
+            p[find(a)] = find(b)
+
+        cnt = 0
+        for i in range(n):
+            if i == find(i):
+                cnt += 1
+        return n - cnt
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    private int[] p;
+
+    public int minSwapsCouples(int[] row) {
+        int n = row.length >> 1;
+        p = new int[n];
+        for (int i = 0; i < n; ++i) {
+            p[i] = i;
+        }
+        for (int i = 0; i < row.length; i += 2) {
+            int a = row[i] >> 1, b = row[i + 1] >> 1;
+            p[find(a)] = find(b);
+        }
+        int cnt = 0;
+        for (int i = 0; i < n; ++i) {
+            if (i == find(i)) {
+                ++cnt;
+            }
+        }
+        return n - cnt;
+    }
+
+    private int find(int x) {
+        if (p[x] != x) {
+            p[x] = find(p[x]);
+        }
+        return p[x];
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> p;
+
+    int minSwapsCouples(vector<int> &row) {
+        int n = row.size() >> 1;
+        p.resize(n);
+        for (int i = 0; i < n; ++i)
+        {
+            p[i] = i;
+        }
+        for (int i = 0; i < row.size(); i += 2)
+        {
+            int a = row[i] >> 1, b = row[i + 1] >> 1;
+            p[find(a)] = find(b);
+        }
+        int cnt = 0;
+        for (int i = 0; i < n; ++i)
+        {
+            if (i == find(i))
+                ++cnt;
+        }
+        return n - cnt;
+    }
+
+    int find(int x) {
+        if (p[x] != x)
+        {
+            p[x] = find(p[x]);
+        }
+        return p[x];
+    }
+};
+```
 
+### **Go**
+
+```go
+var p []int
+
+func minSwapsCouples(row []int) int {
+	n := len(row) >> 1
+	p = make([]int, n)
+	for i := 0; i < n; i++ {
+		p[i] = i
+	}
+	for i := 0; i < len(row); i += 2 {
+		a, b := row[i]>>1, row[i+1]>>1
+		p[find(a)] = find(b)
+	}
+	cnt := 0
+	for i := 0; i < n; i++ {
+		if i == find(i) {
+			cnt++
+		}
+	}
+	return n - cnt
+}
+
+func find(x int) int {
+	if p[x] != x {
+		p[x] = find(p[x])
+	}
+	return p[x]
+}
 ```
 
 ### **...**

solution/0700-0799/0765.Couples Holding Hands/README_EN.md

@@ -6,7 +6,7 @@
 
 <p>
 
-N couples sit in 2N seats arranged in a row and want to hold hands.  We want to know the minimum number of swaps so that every couple is sitting side by side.  A <i>swap</i> consists of choosing <b>any</b> two people, then they stand up and switch seats. 
+N couples sit in 2N seats arranged in a row and want to hold hands. We want to know the minimum number of swaps so that every couple is sitting side by side. A <i>swap</i> consists of choosing <b>any</b> two people, then they stand up and switch seats.
 
 </p><p>
 
@@ -16,8 +16,6 @@ The people and seats are represented by an integer from <code>0</code> to <code>
 
 The couples' initial seating is given by <code>row[i]</code> being the value of the person who is initially sitting in the i-th seat.
 
-
-
 <p><b>Example 1:</b><br /><pre>
 
 <b>Input:</b> row = [0, 2, 1, 3]
@@ -28,8 +26,6 @@ The couples' initial seating is given by <code>row[i]</code> being the value of
 
 </pre></p>
 
-
-
 <p><b>Example 2:</b><br /><pre>
 
 <b>Input:</b> row = [3, 2, 0, 1]
@@ -40,13 +36,11 @@ The couples' initial seating is given by <code>row[i]</code> being the value of
 
 </pre></p>
 
-
-
 <p>
 
 <b>Note:</b>
 
-<ol> 
+<ol>
 
 <li> <code>len(row)</code> is even and in the range of <code>[4, 60]</code>.</li>
 
@@ -61,13 +55,129 @@ The couples' initial seating is given by <code>row[i]</code> being the value of
 ### **Python3**
 
 ```python
-
+class Solution:
+    def minSwapsCouples(self, row: List[int]) -> int:
+        n = len(row) >> 1
+        p = [i for i in range(n)]
+
+        def find(x):
+            if p[x] != x:
+                p[x] = find(p[x])
+            return p[x]
+
+        for i in range(0, len(row), 2):
+            a, b = row[i] >> 1, row[i + 1] >> 1
+            p[find(a)] = find(b)
+
+        cnt = 0
+        for i in range(n):
+            if i == find(i):
+                cnt += 1
+        return n - cnt
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    private int[] p;
+
+    public int minSwapsCouples(int[] row) {
+        int n = row.length >> 1;
+        p = new int[n];
+        for (int i = 0; i < n; ++i) {
+            p[i] = i;
+        }
+        for (int i = 0; i < row.length; i += 2) {
+            int a = row[i] >> 1, b = row[i + 1] >> 1;
+            p[find(a)] = find(b);
+        }
+        int cnt = 0;
+        for (int i = 0; i < n; ++i) {
+            if (i == find(i)) {
+                ++cnt;
+            }
+        }
+        return n - cnt;
+    }
+
+    private int find(int x) {
+        if (p[x] != x) {
+            p[x] = find(p[x]);
+        }
+        return p[x];
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<int> p;
+
+    int minSwapsCouples(vector<int> &row) {
+        int n = row.size() >> 1;
+        p.resize(n);
+        for (int i = 0; i < n; ++i)
+        {
+            p[i] = i;
+        }
+        for (int i = 0; i < row.size(); i += 2)
+        {
+            int a = row[i] >> 1, b = row[i + 1] >> 1;
+            p[find(a)] = find(b);
+        }
+        int cnt = 0;
+        for (int i = 0; i < n; ++i)
+        {
+            if (i == find(i))
+                ++cnt;
+        }
+        return n - cnt;
+    }
+
+    int find(int x) {
+        if (p[x] != x)
+        {
+            p[x] = find(p[x]);
+        }
+        return p[x];
+    }
+};
+```
 
+### **Go**
+
+```go
+var p []int
+
+func minSwapsCouples(row []int) int {
+	n := len(row) >> 1
+	p = make([]int, n)
+	for i := 0; i < n; i++ {
+		p[i] = i
+	}
+	for i := 0; i < len(row); i += 2 {
+		a, b := row[i]>>1, row[i+1]>>1
+		p[find(a)] = find(b)
+	}
+	cnt := 0
+	for i := 0; i < n; i++ {
+		if i == find(i) {
+			cnt++
+		}
+	}
+	return n - cnt
+}
+
+func find(x int) int {
+	if p[x] != x {
+		p[x] = find(p[x])
+	}
+	return p[x]
+}
 ```
 
 ### **...**