From cebf10de23726c44377e9c8ba8957e211dfce98a Mon Sep 17 00:00:00 2001
From: gameloader <ggwqqo@163.com>
Date: Mon, 25 Mar 2024 00:16:14 +0800
Subject: [PATCH] leetcode update

---
 content/posts/leetcode.md | 43 +++++++++++++++++++++++++++++++++++++++
 1 file changed, 43 insertions(+)

diff --git a/content/posts/leetcode.md b/content/posts/leetcode.md
index 320d1b6..2f74579 100644
--- a/content/posts/leetcode.md
+++ b/content/posts/leetcode.md
@@ -1972,3 +1972,46 @@ func reorderList(head *ListNode) {
 	}
 }
 ```
+
+## day27 2024-03-24
+
+### 287. Find the Duplicate Number
+
+Given an array of integers nums containing n + 1 integers where each integer is in the range [1, n] inclusive.
+
+There is only one repeated number in nums, return this repeated number.
+
+You must solve the problem without modifying the array nums and uses only constant extra space.
+
+![0324Ckibjiatyqu7](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0324Ckibjiatyqu7.png)
+
+### 题解
+
+根据鸽笼原理, 显然必有一个数字有一个重复数字. 最简单的思路就是用每个数和数组其余数字比较, 直到找到相同的数为止, 显然这个方法的时间复杂度比较高, 要O(n^2). 最终并没有想出题目中要求的时间复杂度为O(n), 空间复杂度为O(1)的解法, 看题解得知使用的是Floyd 循环检测算法, 这个算法一般用于检测链表中是否存在环, 即使用快慢指针, 同时遍历链表, 如果存在环, 快指针最终会追上慢指针. 如果链表没有环, 则遍历链表最终会到达空指针自然停止, 这里的快慢指针是给了未知长度的链表一个停止条件, 可以避免若链表有环无限循环遍历下去. 这里将数组堪称链表是非常精妙的思路, 将数组中的值看作下一个节点在数组中的下标, 这样如果有两个节点相同, 则最终快指针和慢指针指向的下标会相同, 再从头遍历一次数组, 找到值为这个下标的数据, 即为重复的数.
+
+### 代码
+
+```go
+func findDuplicate(nums []int) int {
+    // Step 1: Find the intersection point of the two pointers
+    slow := nums[0]
+    fast := nums[0]
+
+    for {
+        slow = nums[slow]
+        fast = nums[nums[fast]]
+        if slow == fast {
+            break
+        }
+    }
+
+    // Step 2: Find the entrance to the cycle (duplicate number)
+    slow = nums[0]
+    for slow != fast {
+        slow = nums[slow]
+        fast = nums[fast]
+    }
+
+    return slow
+}
+```