【LeetCode】SingleNumberII_137

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package com.leetCode;/** * 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? *  * @author Zealot * @date 2015年7月25日 下午6:16:12 */public class SingleNumberII_137 {public int singleNumber(int[] nums) {int ones = 0, twos = 0;    for(int i = 0; i < nums.length; i++){        ones = (ones ^ nums[i]) & ~twos;        twos = (twos ^ nums[i]) & ~ones;    }    return ones;}//遍历数组中每一个元素//计算每一个元素中，2进制中每一位加到一个总的数组里边，都加到一起之后，再与3取余，因为除了一个之外，都是出现了3次，只有那一个之出现了不到3次//最后数组转成数字public int singleNumber2(int[] A) {        if(A == null || A.length == 0) return 0;        int[] a = new int[32];        for(int i = 0; i < A.length; i++) {            for(int j = 0; j < 32; j++) {                if((A[i] & (1 << j)) != 0)                    a[j] = (a[j] + 1) % 3;            }        }        int result = 0;        for(int i = 0; i < 32; i++) {            if(a[i] > 0)                result |= (a[i] << i);        }        return result;    }public static void main(String[] args) {SingleNumberII_137 s = new SingleNumberII_137();int[] nums = {1,2,3,3,3,2,5,2,1,1};System.out.println(s.singleNumber2(nums));//System.out.println(77>>1);//System.out.println(4>>1);//System.out.println(Integer.toBinaryString(4).length());//System.out.println("123456".length());}}
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