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Single Number II

时间:2014-07-01 06:23:08      阅读:360      评论:0      收藏:0      [点我收藏+]

题目

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

    public int singleNumber(int[] A) {
        int[] sum = new int[33];
        for (int i = 0; i < A.length; i++) {
        	long  num = A[i];
        	if (num < 0) {
        		sum[32] += 1;
        		num = -num;
        	}
        	int j = 0;
        	while(num > 0) {
        		sum[j] += num % 2;
        		num = num/2;
        		j++;
        	}
        }
        int result = 0;
        for (int i = 31; i >= 0; i--) {
        	sum[i] = sum[i] % 3;
        	if (sum[i] == 0) {
        		result *= 2;
        	} else {
        		result = result * 2 + 1;
        	}
        }
        sum[32] = sum[32] % 3;
        if (sum[32] == 1) {
        	result = -result;
        }
        return result;
    }


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Single Number II

原文:http://blog.csdn.net/u010378705/article/details/35991821

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