来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/find-all-numbers-disappeared-in-an-array著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。 解题思路 哈希表 虽然题目中又说能在不使用额外空间且时间复杂度为O(n)的情况下完成这个任务吗?但是如果我不能,好歹也要告诉面试官能...
}; 在CareerCup中有一道类似的题,5.7 Find Missing Integer 查找丢失的数,但是那道题不让我们直接访问整个int数字,而是只能访问其二进制表示数中的某一位,强行让我们使用位操作Bit Manipulation来解题,也是蛮有意思的一道题。 LeetCode All in One 题目讲解汇总(持续更新中...)...
Runtime:160 ms, faster than30.48% of Python3 online submissions for Missing Number. Memory Usage:15.1 MB, less than39.86% of Python3 online submissions for Missing Number. 【解法二】 思路同上,另一种写法 classSolution:defmissingNumber(self, nums): nums.sort()#Ensure that n is at the last ...
【Leetcode】Missing Number 题目链接:https://leetcode.com/problems/missing-number/ 题目: n distinct numbers taken from0, 1, 2, ..., n, find the one that is missing from the array. For example, Given nums =[0, 1, 3]return2. Note: Your algorithm should run in linear runtime complex...
class Solution: def missingNumber(self, nums: List[int]) -> int: # ans 初始化为 0 ^ n = n ,因为下标的范围为 [0, n - 1] ans: int = len(nums) # 带下标遍历 nums for i, num in enumerate(nums): # 异或下标 ans ^= i # 异或数组内的数 ans ^= num # 此时 ans 就是 [0,...
268. 丢失的数字 - 给定一个包含 [0, n] 中 n 个数的数组 nums ,找出 [0, n] 这个范围内没有出现在数组中的那个数。 示例 1: 输入:nums = [3,0,1] 输出:2 解释:n = 3,因为有 3 个数字,所以所有的数字都在范围 [0,3] 内。2 是丢失的数字,因为它没有出现在 num
LeetCode笔记:268. Missing Number 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 ...
题目地址:https://leetcode.com/problems/missing-number/#/description 题目描述 Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array. ...
missing=4^(0^0)^(1^1)^(2^2)^(4^3) =3^(0^0)^(1^1)^(2^2)^(4^4) =3 其中最开始的4就是数组的长度。 此解法的时间复杂度是O(n),空间复杂度是O(1)。 publicintmissingNumber3(int[]nums){if(nums==null||nums.length<1){return0;}intxor=nums.length;for(inti=0;i<nums.length...
public int missingNumber(int[] nums) { int res = 0; for(int i = 0; i <= nums.length; i++){ res ^= i == nums.length ? i : i ^ nums[i]; } return res; } } First Missing Positive Given an unsorted integer array, find the first missing positive integer. ...