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137.
137.
Single Number II
Givenanarrayofintegers,everyelementappearsthreetimesexceptforone.Findthatsingleone.Note:Youralgorithmshouldhavealinearruntimecomplexity.Couldyouimplementitwithoutusingextramemory?SingleNumberI解法把每个数字转
偏爱纯白色
·
2016-01-31 21:00
[leetcode]
137.
Single Number II 解题报告
题目链接:https://leetcode.com/problems/single-number-ii/Givenanarrayofintegers,everyelementappears three timesexceptforone.Findthatsingleone.Note:Youralgorithmshouldhavealinearruntimecomplexity.Couldyouim
qq508618087
·
2016-01-03 07:00
LeetCode
算法
bit
manipulation
137.
Single Number II
题目: 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 implemen
·
2015-11-13 00:07
number
【LeetCode】
137.
Single Number II (3 solutions)
Single Number II Given an array of integers, every element appears threetimes except for one. Find that single one. Note:Your algorithm should have a linear runtime complexity. Could you implem
·
2015-11-08 11:50
LeetCode
sgu
137.
Funny Strings 线性同余,数论,构造 难度:3
137.
Funny Strings time limit per test: 0.25 sec.
·
2015-10-31 09:37
String
LeetCode
137.
Single Number II
n-1个数字出现3次,1个数字出现1次。位运算,用cnt[32]记录每个位上,二进制'1'出现的次数。1)假设一个数字7,它出现3次,那么二进制0000 0000 0000 0000 0000 0000 00000111会出现三次,也就是第0,1,2位上的二进制'1'会出现3次,那么我们将它%3,得到的余数为0. 2)假设一个数字15,它只出现1次,那么二进制0000 0000 0000 0000
u014674776
·
2015-02-17 11:00
LeetCode
C++
代表不同子网掩码的CIDR值(如192.168.0.0/24)
类地址默认掩码)2.掩码255.128.0.0:/93.掩码255.192.0.0:/104.掩码255.224.0.0:/115.掩码255.240.0.0:/126.掩码255.248.0.0:/
137
wazjajl
·
2013-07-15 17:20
子网掩码
CIDR
Linux系统相关
代表不同子网掩码的CIDR值(如192.168.0.0/24)
类地址默认掩码)2.掩码255.128.0.0:/93.掩码255.192.0.0:/104.掩码255.224.0.0:/115.掩码255.240.0.0:/126.掩码255.248.0.0:/
137
wazjajl
·
2013-07-15 17:20
CIDR
子网掩码
子网划分实例,CIDR 详解
类地址默认掩码)2.掩码255.128.0.0:/93.掩码255.192.0.0:/104.掩码255.224.0.0:/115.掩码255.240.0.0:/126.掩码255.248.0.0:/
137
jafyoo
·
2011-11-25 10:20
职场
详解
CIDR
休闲
子网划分实例
DB2的一些函数
SELECT CORRELATION(SALARY,BONUS) FROM BSEMPMS;
137.
COUNT() 返回一組行或值的個數. SELECT COUNT(*)
menjoy
·
2007-07-31 18:00
db2
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