链表环问题：判断环，求相遇点...

Author: OneCodeMonkey

Tests/CMakeLists.txt

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+/**
+ * AC: Runtime: 0 ms, faster than 100.00% of Java online submissions for Linked List Cycle.
+ * Memory Usage: 39.8 MB, less than 69.19% of Java online submissions for Linked List Cycle.
+ * 
+ * 思路：快慢指针法， T:O(n), S:O(1), n 为链表长度
+ */
+/**
+ * Definition for singly-linked list.
+ * class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) {
+ *         val = x;
+ *         next = null;
+ *     }
+ * }
+ */
+public class Solution {
+    public boolean hasCycle(ListNode head) {
+        // 长度0，1情况
+        if (head == null || head.next == null) {
+            return false;
+        }
+
+        ListNode fasterPointer = head;
+        while (head != null) {
+            head = head.next;
+            if (fasterPointer == null) {
+                return false;
+            }
+            fasterPointer = fasterPointer.next;
+            if (fasterPointer == null) {
+                return false;
+            }
+            fasterPointer = fasterPointer.next;
+            if (head == fasterPointer) {
+                return true;
+            }
+        }
+
+        return false;
+    }
+}

leetcode_solved/[editing]leetcode_0141_Linked_List_Cycle.cpp

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+/**
+ * AC: Runtime: 0 ms, faster than 100.00% of Java online submissions for Linked List Cycle II.
+ *     Memory Usage: 38.9 MB, less than 81.24% of Java online submissions for Linked List Cycle II.
+ * 
+ * 思路：快慢指针法。 细节的是需要证明当快慢两指针相遇时，慢指针再走一倍已走过的路程，最终会停在环处。
+ * 复杂度：T:O(n), S:O(s)
+ */
+/**
+ * Definition for singly-linked list.
+ * class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) {
+ *         val = x;
+ *         next = null;
+ *     }
+ * }
+ */
+public class Solution {
+    public ListNode detectCycle(ListNode head) {
+        // 坑：注意链表 length >= 0，单独处理0的情况，否则 fasterPointer.next 抛空指针异常
+        if (head == null) {
+            return head;
+        }
+
+        ListNode slowerPointer = head, fasterPointer = head;
+        while (fasterPointer.next != null && fasterPointer.next.next != null) {
+            slowerPointer = slowerPointer.next;
+            fasterPointer = fasterPointer.next.next;
+            // 有环
+            // 快慢指针相遇
+            if (slowerPointer == fasterPointer) {
+                // 设置第二个慢指针，从链表头开始。可以证明第二个慢指针与第一个慢指针会在链表的环处相遇。
+                ListNode tempPointer = head;
+                while (slowerPointer != tempPointer) {
+                    slowerPointer = slowerPointer.next;
+                    tempPointer = tempPointer.next;
+                }
+                return slowerPointer;
+            }
+        }
+
+        // 无环
+        return fasterPointer.next == null ? fasterPointer.next : fasterPointer.next.next;
+    }
+}