136. Single Number C++ 答案

136. Single Number -- Easy

解答

相同的数,XOR 等于 0,所以,将所有的数字 XOR 就可以得到只出现一次的数

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int s = 0;

        for(int i = 0; i < nums.size(); i++) {
            s = s ^ nums[i];
        }

        return s;
    }
};

参考

原文地址:https://www.cnblogs.com/optcheng/p/9783186.html

时间: 2024-10-10 06:17:21

136. Single Number C++ 答案的相关文章

LeetCode 136. Single Number &amp; 268. Missing Number

136. Single Number 考察的是异或运算.相同的数异或结果为0,一个数与0异或还是原来的数,以及异或符合交换律.因此,把所有的数都异或起来,结果就是落单的那个数. class Solution { public: int singleNumber(vector<int>& nums) { int res=0; for (int num:nums){ res ^= num; } return res; } }; 268. Missing Number 可以用数学方法直接做,

【59】136. Single Number

136. Single Number Description Submission Solutions Add to List Total Accepted: 191020 Total Submissions: 360448 Difficulty: Easy Contributors: Admin Given an array of integers, every element appears twice except for one. Find that single one. Note:Y

136. Single Number &amp;&amp; 137. Single Number II &amp;&amp; 260. Single Number III

136. Single Number Given an array of integers, every element appears twice except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? Subscribe to see which co

136. Single Number - LeetCode

Question 136.?Single Number Solution 思路:构造一个map,遍历数组记录每个数出现的次数,再遍历map,取出出现次数为1的num public int singleNumber(int[] nums) { Map<Integer, Integer> countMap = new HashMap<>(); for (int i=0; i<nums.length; i++) { Integer count = countMap.get(nums

136.Single Number

Given an array of integers, every element appears twice except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 思路: 一开始自己的思路非常简单,首先设置一个status的变量,然后通过2层for循环

136. Single Number leetcode做题报告

Given an array of integers, every element appears twice except for one. Find that single one.Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? class Solution { public: int singleNumber(vec

LeetCode 136 Single Number(只出现一次的数字)

翻译 给定一个整型数组,除了某个元素外其余元素均出现两次.找出这个只出现一次的元素. 备注: 你的算法应该是一个线性时间复杂度.你可以不用额外空间来实现它吗? 原文 Given an array of integers, every element appears twice except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you i

LeetCode Problem 136:Single Number

描述:Given an array of integers, every element appears twice except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 题目要求O(n)时间复杂度,O(1)空间复杂度. 思路1:初步使用暴力搜索,遍历

LeetCode 136 Single Number(仅仅出现一次的数字)

翻译 给定一个整型数组,除了某个元素外其余元素均出现两次. 找出这个仅仅出现一次的元素. 备注: 你的算法应该是一个线性时间复杂度. 你能够不用额外空间来实现它吗? 原文 Given an array of integers, every element appears twice except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could yo