[leetcode]Single Number II

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?

算法思路：

思路1：用一个数组来模拟CPU位数，记录int类型的32位中每一位出现的次数，容易理解，可惜的是没有达到题中要求的O(1)空间。

public class Solution {
    public int singleNumber(int[] a) {
        int[] binary = new int[32];
        for(int tem : a){
            int bin = 1;
            for(int i = 0; i < 32; i++){
                if((tem & bin) == bin)
                    binary[i]++;
                bin = bin << 1;
            }
        }
        int res = 0;
        for(int i = 0; i < 32; i++){
            res |= (binary[i] % 3) << i;
        }
        return res;
    }
}

思路2：对思路1的优化，对每一位进行数组的遍历，求出该位出现的次数，然后%3取模，对结果的该位进行赋值。

public class Solution {
    public int singleNumber(int[] a) {
        int res = 0;
        for(int i = 0; i < 32; i++){
            int binary = 1 << i;
            int count  = 0;
            for(int tem: a){
                if((tem & binary) != 0){
                    count++;
                } 
            }
            res |= (count % 3) << i;
        }
        return res;
    }
}

 

参考资料：http://www.cnblogs.com/jdflyfly/p/3828929.html