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?
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; }
Single Number II,布布扣,bubuko.com
原文地址:http://blog.csdn.net/u010378705/article/details/35991821