图解面试题：找出数组中重复的数字？

忘是亡心i 2023-10-05 09:33 64阅读 0赞

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今天分享的题目来源于 LeetCode 上的剑指 Offer 系列 **面试题03. 数组中重复的数字**。

题目链接：https://leetcode-cn.com/problems/shu-zu-zhong-zhong-fu-de-shu-zi-lcof/

#### 一、题目描述 ####

找出数组中重复的数字。

在一个长度为 n 的数组 nums 里的所有数字都在 0～n-1 的范围内。数组中某些数字是重复的，但不知道有几个数字重复了，也不知道每个数字重复了几次。请找出数组中任意一个重复的数字。

**示例 1：**

输入：
    [2, 3, 1, 0, 2, 5, 3]
    输出：2 或 3

**限制：**

2 <= n <= 100000

#### 二、题目解析 ####

注意题目描述：一个长度为 n 的数组 nums 里的所有数字都在 0～n-1 的 范围内，这个 **范围** 恰好与数组的下标可以一一对应。

所以我们可以执行某种操作，使索引与值一一对应**，**即索引 0 的值为 0，索引 1 的值为 1。而一旦某个索引的值不只一个，则找到了重复的数字，也即发生了 **哈希冲突**。

#### 三、动画描述 ####

#### 四、图片描述 ####

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#### 五、参考代码    ####

class Solution {
        public int findRepeatNumber(int[] nums) {
            //设索引初始值为 i = 0
            int i = 0;
            //遍历整个数组 nums 
            while(i < nums.length) {
                //索引 i 的值为 i,无需执行交换操作，查看下一位
                if(nums[i] == i) {
                    i++;
                    continue;
                }
                //索引 nums[i] 处的值也为 nums[i]，即找到一组相同值，返回 nums[i] 即可
                if(nums[nums[i]] == nums[i]) return nums[i];
                //执行交换操作，目的是为了使索引与值一一对应，即索引 0 的值为 0，索引 1 的值为 1
                int tmp = nums[i];
                nums[i] = nums[tmp];
                nums[tmp] = tmp;
            }
            //如果遍历整个数组都没有找到相同的值，返回 -1
            return -1;
        }
    }

#### 六、复杂度分析 ####

##### 时间复杂度 #####

遍历数组需要 O(N) 时间。

注意参考代码里面的关键字 **continue**，这表示在 `while` 的一次循环里面，只有这次循环将 **索引(i)** 与 **索引值(num\[i\])** 匹配到了，才会执行下一次循环。

在每一次的循环过程中，**索引(i)** 与 **索引值(num\[i\])** 匹配到后，在后续的循环过程中不会操作它们，所以虽然一开始的循环过程中，执行的交换操作较多，但在后续的循环过程中根本不需要再执行操作了。

根据**均摊复杂度分析** ，总的时间复杂度为 **O(N)** ，N 为数组的长度。

##### 空间复杂度 #####

使用常数复杂度的额外空间，为 **O(1)**。

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