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Single Number II

时间:2014-05-30 15:16:35      阅读:474      评论:0      收藏:0      [点我收藏+]

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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class Solution {
public:
    int singleNumber(int A[], int n) {
        const int LENGTH=32;
        
        int dig[LENGTH];
        for(int i=0;i<LENGTH;i++)
            dig[i]=0;
        
        for(int i=0;i<n;i++)
        {
            unsigned num=A[i];
            for(int j=0;j<LENGTH && num!=0;j++)
            {
                if(num & 0x1==1)
                {
                    dig[j]++;
                    if(dig[j]==3) dig[j]=0;
                }
                num=(num>>1);
            }
        }
        
        unsigned num=0;
        for(int i=0;i<LENGTH;i++)
        {
            num=(dig[i]<<i)+num;
        }
        return num;
    }
};
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Single Number II

原文:http://www.cnblogs.com/erictanghu/p/3759700.html

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