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? 题意:一个整数数组中,除了一个数以外的其他数字都出现了两次…

Leetccode 136 SingleNumber I 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? 1.自己想到了先排序,再遍历数…

/** * Source : https://oj.leetcode.com/problems/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…

SingleNumber I: 题目链接:https://leetcode-cn.com/problems/single-number/ 题意: 给定一个非空整数数组,除了某个元素只出现一次以外,其余每个元素均出现两次.找出那个只出现了一次的元素. 说明: 你的算法应该具有线性时间复杂度. 你可以不使用额外空间来实现吗? 示例 1: 输入: [2,2,1] 输出: 1 示例 2: 输入: [4,1,2,1,2] 输出: 4 分析: 利用异或(xor)运算,两个相同的数异或为0 代码如下: cla…

LeetCode--single-number系列 Question 1 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? Soluti…

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? //意思是一堆数中找出没成对的那个,用异或效率最高A^0=A   A^A=0 最后可求…

题目意思: 一个数值数组中,大部分的数值出现两次,仅仅有一个数值仅仅出现过一次,求编程求出该数字. 要求,时间复杂度为线性,空间复杂度为O(1). 解题思路: 1.先排序.后查找. 因为排序的最快时间是O(nlogn), 所以这样的方法不能满足时间的要求. 2.其他技巧来解决: 依据主要的计算机组成原理的知识.採用"异或运算"来巧妙的完毕. 异或运算非常easy: 0^0=0 1^1=0 1^0=0^1=1 也就是说同样则为0,不同则为1. 所以不论什么同样的两个数的异或运算值结果为0…

题目:singleNumber 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? 给一个数组里面除了唯一的一个数只出现一次,其他每个数都…

Given an array of integers, every element appears three times except for one. Find that single one. solution: def singleNumber(self,A): A.sort() for i in range(0,len(A)): if i%3=0 && A[i]!=A[i+1] && A[i]!=A[i+2kjk]: return A[i-1] if __name…

Given an array of integers, every element appears twice except for one. Find that single one. solution: class solution: def singleNumber(self,A): A.sort() for i in range(1,len(A)): if i%2=1 && A[i]!=A[i-1]: return A[i-1] if __name__ == '__main__':…

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? C++ class Solution { public: int singleNumb…

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. For example: Given nums = [1, 2, 1, 3, 2, 5], return [3, 5]. Note: The order…

Given an array of integers, every element appears three times except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? 这道题是之前那道 Single Number 单独的数字的延伸,那道题的解法…

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? 本来是一道非常简单的题,但是由于加上了时间复杂度必须是O(n),并且空间复杂度为O(1)…

Array 448.找出数组中所有消失的数 要求:整型数组取值为 1 ≤ a[i] ≤ n,n是数组大小,一些元素重复出现,找出[1,n]中没出现的数,实现时时间复杂度为O(n),并不占额外空间 思路1:(discuss)用数组下标标记未出现的数,如出现4就把a[3]的数变成负数,当查找时判断a的正负就能获取下标 tips:注意数组溢出 public List<Integer> findDisappearedNumbers(int[] nums) { List<Integer> d…

82. Remove Duplicates from Sorted List II https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/ 下面代码是自写hashmap版,练一下hashmap,用一般的map<int,int>红黑树map也能过. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *…

利用堆栈:http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/http://oj.leetcode.com/problems/longest-valid-parentheses/ (也可以用一维数组,贪心)http://oj.leetcode.com/problems/valid-parentheses/http://oj.leetcode.com/problems/largest-rectangle-in-histo…

(记忆线:当时一刷完是1-205. 二刷88道.下次更新记得标记不能bug-free的原因.)   88-------------Perfect Squares(完美平方数.给一个整数,求出用平方数来相加得到最小的个数) public class Solution{ public static void main(String[] args){ System.out.println(numSquares(8)); } public static int numSquares(int n){ //…

-------------------------------- 这道题好坑啊,自己说是2*n+1个数字,结果有组测试数据竟然传了个空数组进来... 经典位算法: n^n==0 n^0==n AC代码: public class Solution { /** *@param A : an integer array *return : a integer */ public int singleNumber(int[] A) { if(A.length==0) return 0; int n=A…

最原始的方法:先排序,然后从头查找.若nums[i] = nums[i] + 1则为一对相同的数,i = i  + 2,继续判断.若nums[i] != nums[i] + 1,则输出nums[i].需注意不要越界! class Solution { public: int singleNumber(vector<int>& nums) { int t; ) { t = nums[]; } else { sort(nums.begin(), nums.end()); ; i <…

以下列举一些在开发中可能用得上的UI控件: IBAction和IBOutlet,UIView 1 @interface ViewController : UIViewController 2 3 @property(nonatomic, weak)IBOutlet UILabel *lable; 4 5 @end 6 7 8 9 @interface ViewController () 10 11 @end 12 13 @implementation ViewController 14 15 /…

题目1 落单的数 给出2*n + 1 个的数字,除其中一个数字之外其他每个数字均出现两次,找到这个数字. 链接:http://www.lintcode.com/zh-cn/problem/single-number/ 样例 给出 [1,2,2,1,3,4,3],返回 4 挑战 一次遍历,常数级的额外空间复杂度 解决方案 方法1思路:将所有的数转换成二进制,因为是int类型,共32位.申请常数级(32位)的额外空间,然后每个数对应的位相加,最后对应位上的和模2.最后的结果就是单个数对应的二进制数.…

[1]LeetCode 136 Single Number 题意:奇数个数,其中除了一个数只出现一次外,其他数都是成对出现,比如1,2,2,3,3...,求出该单个数. 解法:容易想到异或的性质,两个相同的数异或为0,那么把这串数从头到尾异或起来,最后的数就是要求的那个数. 代码如下: class Solution { public: int singleNumber(vector<int>& nums) { ; ;i<nums.size();i++) sum ^= nums[i…

-------------------------------------- 一个数异或它自己会得到0,0异或n会得到n,所以可以用异或来消除重复项. AC代码如下: public class Solution { public int singleNumber(int[] nums) { int res=0; for(Integer i:nums) res^=i; return res; } } 题目来源: https://leetcode.com/problems/single-number…

问题: Given an array of integers, every element appears three times except for one. Find that single one. Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?   Single Number I 升级版,一个数组中其它数出现了…

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? 感谢cnblog博主feiling,这篇博客一方面参考了feiling的博客,也加入了…

Given 2*n + 1 numbers, every numbers occurs twice except one, find it. Have you met this question in a real interview? Yes Example Given [1,2,2,1,3,4,3], return 4 Challenge One-pass, constant extra space. LeetCode上的原题,请参见我之前的博客Single Number. class So…

463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime…

Single Number I : 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? Solution: 解法不少,贴一种: class…

Given an array of integers, every element appears twice except for one. Find that single one. 思路: 最经典的方法,利用两个相同的数异或结果为0的性质,则将整个数组进行异或,相同的数俩俩异或,最后得到的就是那个single number,复杂度是O(n) 代码: class Solution { public: int singleNumber(vector<int>& nums) { int…