增加了lcof-38.52和lc-1195.1226的题解 (#341)

ChunelFeng · junfeng.fj · web-flow · commit cf3d2c676330 · 2021-01-31T18:44:42.000+08:00
Co-authored-by: junfeng.fj &lt;junfeng.fj@alibaba-inc.com&gt;
diff --git a/lcof/面试题38. 字符串的排列/README.md b/lcof/面试题38. 字符串的排列/README.md
@@ -117,6 +117,40 @@ var permutation = function (s) {
 };
 ```
 
+### **C++**
+
+```cpp
+class Solution {
+public:    
+    void func(string str, int index, set<string>& mySet) {
+        if (index == str.size()) {
+            // 当轮训到最后一个字符的时候，直接放入set中。加入set结构，是为了避免插入的值重复
+            mySet.insert(str);
+        } else {
+            for (int i = index; i < str.size(); i++) {
+                // 从传入位置(index)开始算，固定第一个字符，然后后面的字符依次跟index位置交换
+                swap(str[i], str[index]);
+                int temp = index + 1;
+                func(str, temp, mySet);
+                swap(str[i], str[index]);
+            } 
+        }
+    }
+
+    vector<string> permutation(string s) {
+        set<string> mySet;
+        func(s, 0, mySet);
+        vector<string> ret;
+        for (auto& x : mySet) {
+            /* 这一题加入mySet是为了进行结果的去重。
+               但由于在最后加入了将set转vector的过程，所以时间复杂度稍高 */ 
+            ret.push_back(x);
+        }
+        return ret;
+    }
+};
+```
+
 ### **...**
 
 ```
diff --git a/lcof/面试题38. 字符串的排列/Solution.cpp b/lcof/面试题38. 字符串的排列/Solution.cpp
@@ -0,0 +1,25 @@
+class Solution {
+public:    
+    void func(string str, int index, set<string>& mySet) {
+        if (index == str.size()) {
+            mySet.insert(str);
+        } else {
+            for (int i = index; i < str.size(); i++) {
+                swap(str[i], str[index]);
+                int temp = index + 1;
+                func(str, temp, mySet);
+                swap(str[i], str[index]);
+            } 
+        }
+    }
+
+    vector<string> permutation(string s) {
+        set<string> mySet;
+        func(s, 0, mySet);
+        vector<string> ret;
+        for (string x : mySet) {
+            ret.push_back(x);
+        }
+        return ret;
+    }
+};
diff --git a/lcof/面试题52. 两个链表的第一个公共节点/README.md b/lcof/面试题52. 两个链表的第一个公共节点/README.md
@@ -169,6 +169,36 @@ var getIntersectionNode = function (headA, headB) {
 };
 ```
 
+### **C++**
+
+```cpp
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+ 
+class Solution {
+public:
+    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
+        ListNode* a = headA;
+        ListNode* b = headB;
+        while (a != b) {
+            /* 这个循环的思路是，a先从listA往后走，如果到最后，就接着从listB走；b正好相反。
+               如果有交集的话，a和b会在分别进入listB和listA之后的循环中项目
+               如果没有交集的话，则a和b会同时遍历完listA和listB后，值同时为nullptr */
+            a = (a == nullptr) ? headB : a->next;
+            b = (b == nullptr) ? headA : b->next;
+        }
+
+        return a;
+    }
+};
+```
+
 ### **...**
 
 ```
diff --git a/lcof/面试题52. 两个链表的第一个公共节点/Solution.cpp b/lcof/面试题52. 两个链表的第一个公共节点/Solution.cpp
@@ -0,0 +1,22 @@
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode(int x) : val(x), next(NULL) {}
+ * };
+ */
+ 
+class Solution {
+public:
+    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
+        ListNode* a = headA;
+        ListNode* b = headB;
+        while (a != b) {
+            a = (a == nullptr) ? headB : a->next;
+            b = (b == nullptr) ? headA : b->next;
+        }
+
+        return a;
+    }
+};
diff --git a/solution/1100-1199/1195.Fizz Buzz Multithreaded/README.md b/solution/1100-1199/1195.Fizz Buzz Multithreaded/README.md
@@ -56,6 +56,65 @@
 
 ```
 
+### **C++**
+
+```cpp
+class FizzBuzz {
+private:
+    std::mutex mtx;
+    atomic<int> index;
+    int n;
+
+// 这里主要运用到了C++11中的RAII锁(lock_guard)的知识。
+// 需要强调的一点是，在进入循环后，要时刻不忘加入index <= n的逻辑
+public:
+    FizzBuzz(int n) {
+        this->n = n;
+        index = 1;
+    }
+
+    void fizz(function<void()> printFizz) {
+        while (index <= n) {
+            std::lock_guard<std::mutex> lk(mtx);
+            if (0 == index % 3 && 0 != index % 5 && index <= n) {
+                printFizz();
+                index++;
+            }
+        }
+    }
+
+    void buzz(function<void()> printBuzz) {
+        while (index <= n) {
+            std::lock_guard<std::mutex> lk(mtx);
+            if (0 == index % 5 && 0 != index % 3 && index <= n){
+                printBuzz();
+                index++;
+            }
+        }
+    }
+
+    void fizzbuzz(function<void()> printFizzBuzz) {
+        while (index <= n) {
+            std::lock_guard<std::mutex> lk(mtx);
+            if (0 == index % 15 && index <= n) {
+                printFizzBuzz();
+                index++;
+            } 
+        }
+    }
+
+    void number(function<void(int)> printNumber) {
+        while (index <= n) { 
+            std::lock_guard<std::mutex> lk(mtx);
+            if (0 != index % 3 && 0 != index % 5 && index <= n) {
+                printNumber(index);
+                index++;
+            }
+        }
+    }
+};
+```
+
 ### **...**
 
 ```
diff --git a/solution/1100-1199/1195.Fizz Buzz Multithreaded/Solution.cpp b/solution/1100-1199/1195.Fizz Buzz Multithreaded/Solution.cpp
@@ -0,0 +1,56 @@
+class FizzBuzz {
+private:
+    std::mutex mtx;
+    atomic<int> index;
+    int n;
+
+public:
+    FizzBuzz(int n) {
+        this->n = n;
+        index = 1;
+    }
+
+    // printFizz() outputs "fizz".
+    void fizz(function<void()> printFizz) {
+        while (index <= n) {
+            std::lock_guard<std::mutex> lk(mtx);
+            if (0 == index % 3 && 0 != index % 5 && index <= n) {
+                printFizz();
+                index++;
+            }
+        }
+    }
+
+    // printBuzz() outputs "buzz".
+    void buzz(function<void()> printBuzz) {
+        while (index <= n) {
+            std::lock_guard<std::mutex> lk(mtx);
+            if (0 == index % 5 && 0 != index % 3 && index <= n){
+                printBuzz();
+                index++;
+            }
+        }
+    }
+
+    // printFizzBuzz() outputs "fizzbuzz".
+    void fizzbuzz(function<void()> printFizzBuzz) {
+        while (index <= n) {
+            std::lock_guard<std::mutex> lk(mtx);
+            if (0 == index % 15 && index <= n) {
+                printFizzBuzz();
+                index++;
+            } 
+        }
+    }
+
+    // printNumber(x) outputs "x", where x is an integer.
+    void number(function<void(int)> printNumber) {
+        while (index <= n) { 
+            std::lock_guard<std::mutex> lk(mtx);
+            if (0 != index % 3 && 0 != index % 5 && index <= n) {
+                printNumber(index);
+                index++;
+            }
+        }
+    }
+};
diff --git a/solution/1200-1299/1226.The Dining Philosophers/README.md b/solution/1200-1299/1226.The Dining Philosophers/README.md
@@ -77,6 +77,30 @@ output[i] = [a, b, c] (3个整数)
 
 ```
 
+### **C++**
+
+```cpp
+class DiningPhilosophers {
+public:
+    using Act = function<void()>;
+
+    void wantsToEat(int philosopher, Act pickLeftFork, Act pickRightFork, Act eat, Act putLeftFork, Act putRightFork) {
+		/* 这一题实际上是用到了C++17中的scoped_lock知识。
+		   作用是传入scoped_lock(mtx1, mtx2)两个锁，然后在作用范围内，依次顺序上锁mtx1和mtx2；然后在作用范围结束时，再反续解锁mtx2和mtx1。
+		   从而保证了philosopher1有动作的时候，philosopher2无法操作；但是philosopher3和philosopher4不受影响 */
+        std::scoped_lock lock(mutexes_[philosopher], mutexes_[philosopher >= 4 ? 0 : philosopher + 1]);
+        pickLeftFork();
+        pickRightFork();
+        eat();
+        putLeftFork();
+        putRightFork();
+    }
+
+private:
+    vector<mutex> mutexes_ = vector<mutex>(5);
+};
+```
+
 ### **...**
 
 ```
diff --git a/solution/1200-1299/1226.The Dining Philosophers/Solution.cpp b/solution/1200-1299/1226.The Dining Philosophers/Solution.cpp
@@ -0,0 +1,16 @@
+class DiningPhilosophers {
+public:
+    using Act = function<void()>;
+
+    void wantsToEat(int philosopher, Act pickLeftFork, Act pickRightFork, Act eat, Act putLeftFork, Act putRightFork) {
+        std::scoped_lock lock(mutexes_[philosopher], mutexes_[philosopher >= 4 ? 0 : philosopher + 1]);
+        pickLeftFork();
+        pickRightFork();
+        eat();
+        putLeftFork();
+        putRightFork();
+    }
+
+private:
+    vector<mutex> mutexes_ = vector<mutex>(5);
+};
