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Given an array of integers, every element appears three times except for one. Find that single one.
/** * 将整数所有位相加对3取余,剩下的就是只出现一次的数 * @param A * @return */ public int singleNumber(int[] A) { int sum[] = new int[32]; //保存整数32位,A中所有数每一位相加的和 int res = 0; //出现一次的数 for(int i = 0; i < 32; i++){ for(int j = 0; j < A.length; j++){ sum[i] += ((A[j] >>> i) & 1); }//for sum[i] %= 3; res += (sum[i] << i); }//for return res; }
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原文地址:http://www.cnblogs.com/luckygxf/p/4389022.html