diff --git a/include/s0142_linked_list_cycle.hpp b/include/s0142_linked_list_cycle.hpp
new file mode 100644
index 0000000..d80835d
--- /dev/null
+++ b/include/s0142_linked_list_cycle.hpp
@@ -0,0 +1,24 @@
+#ifndef S0142_LINKED_LIST_CYCLE_HPP
+#define S0142_LINKED_LIST_CYCLE_HPP
+
+#include <unordered_map>
+
+struct ListNode {
+  int val;
+  ListNode* next;
+  ListNode(int x) : val(x), next(nullptr) {}
+};
+
+struct IterResult {
+  ListNode *node;
+  bool meet;
+};
+
+class S0142 {
+ public:
+  std::unordered_map<ListNode *, int> footprint;
+  IterResult iter(ListNode *fast, ListNode* slow, bool startUp);
+  ListNode *detectCycle(ListNode *head);
+};
+
+#endif
diff --git a/notes/src/SUMMARY.md b/notes/src/SUMMARY.md
index 3241eec..9e9f086 100644
--- a/notes/src/SUMMARY.md
+++ b/notes/src/SUMMARY.md
@@ -10,6 +10,7 @@
 # 链表
 
 - [总结](./linked-list.md)
+- [环形链表](./linked_list_cycle.md)
 
 # 字符串
 
diff --git a/notes/src/linked-list.md b/notes/src/linked-list.md
index 7f66395..d59e3d5 100644
--- a/notes/src/linked-list.md
+++ b/notes/src/linked-list.md
@@ -8,14 +8,18 @@
 
 ```cpp
 void iter(ListNode *node) {
+  // 终止条件
   if (node == nullptr) {
     return;
   }
   /*
-   your
-   condition
+   从前往后遍历
    */
   iter(node->next);
+  /*
+   从后往前遍历
+   */
+  return;
 }
 ```
 
diff --git a/notes/src/linked_list_cycle.md b/notes/src/linked_list_cycle.md
new file mode 100644
index 0000000..7b0e803
--- /dev/null
+++ b/notes/src/linked_list_cycle.md
@@ -0,0 +1,40 @@
+# 环形链表
+
+[Leetcode](https://leetcode.com/problems/linked-list-cycle-ii/)
+
+可以用回溯法解这道题。
+
+首先递归遍历链表的一般结构如下：
+
+```cpp
+void iter(ListNode *node) {
+  // 终止条件
+  if (node == nullptr) {
+    return;
+  }
+  /*
+   从前往后遍历
+   */
+  iter(node->next);
+  /*
+   从后往前遍历
+   */
+  return;
+}
+```
+
+而查找链表中是否有环的思路是快慢指针，如果相遇则说明有环。
+
+所以我们可以这样：
+
+终止条件是快指针走到链表末尾或者快慢指针相遇。
+
+从前往后遍历不需要做什么额外操作。
+
+从后往前遍历的时候先把快指针的足迹记录到一个哈希表中，键值是节点地址，值是快指针经过的次数。
+
+当出现以下两种情况的时候就说明找到了环的入口：
+
+1. footprint[fast] == 1 && footprint[fast->next] > 1
+2. footprint[fast->next] == 1 && footprint[fast->next->next] > 1
+
diff --git a/src/s0142_linked_list_cycle.cpp b/src/s0142_linked_list_cycle.cpp
new file mode 100644
index 0000000..0194f49
--- /dev/null
+++ b/src/s0142_linked_list_cycle.cpp
@@ -0,0 +1,70 @@
+#include "s0142_linked_list_cycle.hpp"
+
+IterResult S0142::iter(ListNode *fast, ListNode *slow, bool startUp) {
+  // 终止条件
+  if (fast == nullptr || slow == nullptr) {
+    return {nullptr, false};
+  }
+  if (fast->next == nullptr) {
+    return {nullptr, false};
+  }
+  if (slow->next == slow) {
+    return {slow, true};
+  }
+  // 当两指针相遇时同样达到终止条件
+  if (fast == slow && !startUp) {
+    return {nullptr, true};
+  }
+  // 递归
+  IterResult result = iter(fast->next->next, slow->next, false);
+  // 从后往前遍历
+  // 达到了终止条件，但是没有两个指针遇见过
+  // 这说明没有环
+  if (!result.meet) {
+    return {nullptr, false};
+  } else {
+    // 有环的情况
+    // 我们把 fast 和 fast-next 节点所经过的足迹全部记录下来
+    // key 是所经过的节点
+    // value 是所经过的次数
+    if (footprint.count(fast) == 0) {
+      footprint[fast] = 1;
+    } else {
+      footprint[fast] += 1;
+    }
+    if (footprint.count(fast->next) == 0) {
+      footprint[fast->next] = 1;
+    } else {
+      footprint[fast->next] += 1;
+    }
+    // 什么情况下是环的入口呢？
+    // 1. footprint[fast] == 1 && footprint[fast->next] > 1
+    // 2. footprint[fast->next] == 1 && footprint[fast->next->next] > 1
+    if (footprint.count(fast) == 1 && footprint.count(fast->next) == 1) {
+      if (footprint[fast] == 1 && footprint[fast->next] > 1) {
+        return {fast->next, true};
+      }
+    }
+    if (footprint.count(fast->next) == 1 &&
+        footprint.count(fast->next->next) == 1) {
+      if (footprint[fast->next] == 1 && footprint[fast->next->next] > 1) {
+        return {fast->next->next, true};
+      }
+    }
+    // 返回
+    return result;
+  }
+}
+
+ListNode *S0142::detectCycle(ListNode *head) {
+  if (head == nullptr) {
+    return nullptr;
+  }
+  if (head->next == nullptr) {
+    return nullptr;
+  }
+  // 递归函数无法处理所有节点都在环里的情况，所以在这里加一个哑节点
+  ListNode dummy = ListNode(0);
+  dummy.next = head;
+  return iter(&dummy, &dummy, true).node;
+}
diff --git a/tests/s0142_linked_list_cycle.cpp b/tests/s0142_linked_list_cycle.cpp
new file mode 100644
index 0000000..6f5c1c3
--- /dev/null
+++ b/tests/s0142_linked_list_cycle.cpp
@@ -0,0 +1,57 @@
+#include "s0142_linked_list_cycle.hpp"
+
+#include <gtest/gtest.h>
+
+TEST(Problem142, Case1) {
+  ListNode node1 = ListNode(3);
+  ListNode node2 = ListNode(2);
+  ListNode node3 = ListNode(0);
+  ListNode node4 = ListNode(-4);
+  node1.next = &node2;
+  node2.next = &node3;
+  node3.next = &node4;
+  node4.next = &node2;
+  S0142 solution;
+  EXPECT_EQ(solution.detectCycle(&node1), &node2);
+}
+
+TEST(Problem142, Case2) {
+  ListNode node1 = ListNode(3);
+  ListNode node2 = ListNode(2);
+  ListNode node3 = ListNode(0);
+  ListNode node4 = ListNode(-4);
+  node1.next = &node2;
+  node2.next = &node3;
+  node3.next = &node4;
+  node4.next = &node1;
+  S0142 solution;
+  EXPECT_EQ(solution.detectCycle(&node1), &node1);
+}
+
+TEST(Problem142, Case3) {
+  ListNode node1 = ListNode(3);
+  ListNode node2 = ListNode(2);
+  ListNode node3 = ListNode(0);
+  ListNode node4 = ListNode(-4);
+  node1.next = &node2;
+  node2.next = &node3;
+  node3.next = &node4;
+  S0142 solution;
+  EXPECT_EQ(solution.detectCycle(&node1), nullptr);
+}
+
+TEST(Problem142, Case4) {
+  ListNode node1 = ListNode(1);
+  ListNode node2 = ListNode(2);
+  node1.next = &node2;
+  node2.next = &node1;
+  S0142 solution;
+  EXPECT_EQ(solution.detectCycle(&node1), &node1);
+}
+
+TEST(Problem142, Case5) {
+  ListNode node1 = ListNode(1);
+  node1.next = &node1;
+  S0142 solution;
+  EXPECT_EQ(solution.detectCycle(&node1), &node1);
+}