4.Single Number && Single Number （II）

Single Number：

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?

代码：

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class Solution { public:     int singleNumber(int A[], int n) {         int result = 0;         for(int i = 0; i < n; ++i){             result ^= A[i];         }         return result;     } };

 

Single Number （II）：

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?

 方法1：（分别计算各位1的个数 mod 3）

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class Solution { public:     int singleNumber(int A[], int n) {         int count[32] = {0};         int result = 0;         int bit = sizeof(int) * 8;         for(int i = 0; i < n; ++i) {             for(int j = 0; j < bit; ++j){                 if((A[i] >> j) & 0x1) count[j] = (count[j] + 1) % 3;             }         }         for(int i = 0; i < bit; ++i){             result |= (count[i] << i);         }         return result;     } };

 方法2:(三个变量，分别记录出现一次，出现两次，出现三次的值)

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class Solution { public:     int singleNumber(int A[], int n) {         int one, two, three;         one = two = three = 0;         for(int i = 0; i < n; ++i){             two |= (one & A[i]);             one ^= A[i];             three = ~(one & two);             one &= three;             two &= three;         }         return one;     } };