[leetcode] Single Number II

Single Number II

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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?

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举个例子说明：




如上图所示，用ones记录，出现所有int，响应的位1出现的次数(mod 3)等于1
tows记录，响应的位1出现的次数(mod 3)等于2

class Solution {public:int singleNumber(int A[], int n){//本程序可以用下面的例子说明：//输入A[4] = {3, 3, 3, 1};/*初始化：one = 0; tows = 0; xthrees = 0;开始循环 i = 0tows = tows | (ones & 3) = 0 | (0 & 3) = 0;ones = ones ^ 3 = 0 ^ 0x0000 0011 = 0x0000 0011 = 3;后面三行代码将ones和tows中的同时出现的1去掉ones = 3; tows = 0;第二遍循环i=1tows = tows | (3 & 3) = 0 | (3) = 3;ones = ones ^ 3 = 0;去掉共同的1, ones = 0, tows = 3;第三遍循环i=2tows = tows | (0 & 3) = 3;ones = ones ^ 3 = 3;去掉共同的1,ones = 0, tows = 0;第四遍循环i=3tows = tows | (ones & 1) = 0;ones = ones ^ 1 = 1;去掉共同的1 ones = 1, tows = 0;最终的结果是ones = 1就是只出现一次的那个数*///用ones记录到当前计算的变量为止，二进制1出现“1次”（mod 3 之后的 1）的数位。//举个例子int ones = 0, twos = 0, xthrees = 0;for (int i = 0; i < n; ++i){twos |= (ones & A[i]);ones ^= A[i];//异或操作//下面三行代码的作用是：//将ones和twos中相同位置“都出现”的1设置为0，举例说明：//ones = 0x0010, tows = 0x0011//xthrees = ~(ones&0x11) = ~(0x0010) = 0x1101//ones = one & xthrees = 0x0010 & 0x1101 = 0x0000;//twos = twos & xthrees = 0x0011 & 0x1101 = 0x0001//为啥将ones tows中的相同位置中都出现的1设置为0，是为了//ones和tows都出现，说明，一个数出现了三次，就去掉xthrees = ~(ones & twos);ones &= xthrees;twos &= xthrees;}return ones;}};