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+84,6 @@ # 1. 前言 -本文的绘图可通过以下途径免费获得并使用: - -- [ProcessOn](https://www.processon.com/view/5a3e4c7be4b0909c1aa18b49) -- [DrawIO](https://drive.google.com/file/d/1nSSCpPUC05MFoeFuf_aeTtkm7dG5-bJ1/view?usp=sharing) - 本文内容可在微信小程序中阅读:

@@ -115,11 +110,13 @@ Output: ## 解题思路 -要求复杂度为 O(N) + O(1),也就是时间复杂度 O(N),空间复杂度 O(1)。因此不能使用排序的方法,也不能使用额外的标记数组。牛客网讨论区这一题的首票答案使用 nums[i] + length 来将元素标记,这么做会有加法溢出问题。 +要求复杂度为 O(N) + O(1),也就是时间复杂度 O(N),空间复杂度 O(1)。因此不能使用排序的方法,也不能使用额外的标记数组。 -这种数组元素在 [0, n-1] 范围内的问题,可以将值为 i 的元素调整到第 i 个位置上。 +对于这种数组元素在 [0, n-1] 范围内的问题,可以将值为 i 的元素调整到第 i 个位置上进行求解。 -以 (2, 3, 1, 0, 2, 5) 为例: +以 (2, 3, 1, 0, 2, 5) 为例,遍历到位置 4 时,该位置上的数为 2,但是第 2 个位置上已经有一个 2 的值了,因此可以知道 2 重复: + +

```text position-0 : (2,3,1,0,2,5) // 2 <-> 1 @@ -132,8 +129,6 @@ position-3 : (0,1,2,3,2,5) // already in position position-4 : (0,1,2,3,2,5) // nums[i] == nums[nums[i]], exit ``` -遍历到位置 4 时,该位置上的数为 2,但是第 2 个位置上已经有一个 2 的值了,因此可以知道 2 重复。 - ```java public boolean duplicate(int[] nums, int length, int[] duplication) { if (nums == null || length <= 0)