题目描述:
给定一个数组,里面除了一个数字,其他的都出现三次。求出这个数字
原文描述:
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?
Subscribe to see which companies asked this question
思路:
- 设置一个32位的数组,然然后对数组,Array【i】代表i位的1个数,统计整个数组nums【】的数字,对于每个Array【i】做mod3的运算
- 遍历nums【】数组时候,是左移&1取出,加到Array【i】上面,然后mod3
- 最后右移加到result上面
代码:
// Single Number II
// 方法1,时间复杂度O(n),空间复杂度O(1)
public class Solution {
public int singleNumber(int[] nums) {
final int W = Integer.SIZE; // 一个整数的bit数,即整数字长
int[] count = new int[W]; // count[i]表示在在i位出现的1的次数
for (int i = 0; i < nums.length; i++) {
for (int j = 0; j < W; j++) {
count[j] += (nums[i] >>> j) & 1;
count[j] %= 3;
}
}
int result = 0;
for (int i = 0; i < W; i++) {
result += (count[i] << i);
}
return result;
}
};