[algorithm] sword finger offer2 golang interview question 4: numbers that appear only once

[ Algorithm ] The finger of the sword offer2 golang Interview questions 4： A number that appears only once

subject 1:

Give you an array of integers nums , Except that one element only appears once Outside , Every other element just happens to appear Three times . Please find and return the element that only appears once .

Example 1：

Input ：nums = [2,2,3,2]
Output ：3
Example 2：

Input ：nums = [0,1,0,1,0,1,100]
Output ：100

Tips ：

1 <= nums.length <= 3 * 104
-231 <= nums[i] <= 231 - 1
nums in , Except that one element only appears once Outside , Every other element just happens to appear Three times

Advanced ： Your algorithm should have linear time complexity . Can you do this without using extra space ？

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/WGki4K
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Ideas 1:

// Ideas :  appear 3 Time , Use  bs []int  To record  32  Single digit  1  The number of 
// If   bs[i]  The number of is  0  perhaps  3  Multiple , let  bs[i] = 0, conversely  bs[i] =1
// Final material  bs => int

Code

func singleNumber(nums []int) int {
    
    // Ideas :  appear 3 Time , Use  bs []int  To record  32  Single digit  1  The number of 
    // If  bs[i]  The number of is  0  perhaps  3  Multiple , let  bs[i] = 0, conversely  bs[i] =1
    // Final material  bs => int

    // Processing parameters 
    if len(nums) == 0 {
    
        return 0 
    }

    // structure  bs
    bs := make([]int,32)

    // Traverse element to  bs  assignment 
    for i := 0; i < len(nums); i++ {
    
        for j := 0; j < 32; j++ {
    
            if (nums[i] & (1<<j)) != 0 {
    
                bs[j] ++
            }
        } 
    }
    
    //bs[i]  The number of is  0  perhaps  3  Multiple , let  bs[i] = 0, conversely  bs[i] =1
    for j := 0; j < 32; j++ {
    
        if bs[j]%3 != 1 {
    
            bs[j] = 0 
        }else {
    
            bs[j] = 1
        }
    } 

    // format  bs  by  res int
    //tips: bug golang 64  Bit machine  int  Namely  64  position 
    var res int32
    for j := 0; j < 32; j++ {
    
        if bs[j] == 1 {
    
            res = res | (1 << j)
        }
    } 
    return int(res)
}

test