268. Missing Number

最新推荐文章于 2025-09-13 15:50:46 发布

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原文链接：http://www.cnblogs.com/joycelee/p/5407726.html

文章标签：

#runtime

本文介绍了一个利用位运算快速找到数组中缺失元素的算法，通过异或操作在O(n)时间复杂度和O(1)空间复杂度内解决问题。

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

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287. Find the Duplicate Number

 1 public class Solution {
 2     public int missingNumber(int[] nums) {
 3         int xor = 0;
 4         for (int i = 0; i < nums.length; i++) {
 5             xor ^= i ^ nums[i];
 6         }
 7         xor ^= nums.length;
 8         return xor;
 9     }
10 }