Single Number II
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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?
- //java里int始终占4个字节,32位,我们外层循环遍历32次,然后内层循环记录0-31位每一位出现的次数,
- //内层循环结束后将结果取余于3即为当前位的值
- //时间复杂度O(32 * n), 空间复杂度O(1)
- // 比方说
- //1101
- //1101
- //1101
- //0011
- //0011
- //0011
- //1010 这个unique的
- //----
- //4340 1的出现次数
- //1010 余3的话 就是那个唯一的数!
- public class Solution {
- public int singleNumber(int[] A) {
- int res=0;
- int bit;
- for(int j=0;j<32;j++){
- bit=0;
- for(int i=0;i<A.length;i++){
- if((A[i]>>j&1)==1){
- bit++;
- }
- }
- bit=bit%3;
- res+=(1<<j)*bit;
- }
- return res;
- }
- }
0 0
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