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137. Single Number II
Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
Example 1:
Input: [2,2,3,2] Output: 3
Example 2:
Input: [0,1,0,1,0,1,99] Output: 99
思路分析
由于除了唯一数外，其余数字都出现了三次，那么将每个数字的每位相加，除以 3，余数即为唯一数。
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-09-19 14:48:40
*/
public int singleNumber(int[] nums) {
int[] bits = new int[32];
for (int i = 0; i < 32; i++) {
int cnt = 0;
for (int j = 0; j < nums.length; j++) {
int num = nums[j];
cnt += (num & 1);
nums[j] = (num >>> 1);
}
bits[i] = cnt % 3;
}
int result = 0;
for (int i = 31; i >= 0; i--) {
result <<= 1;
result |= bits[i];
}
return result;
}