【leetcode】【137】Single Number II

1.问题描述

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?

2.问题分析

这个题与Single Number 是一个系列的，都是进行的位运算。但是问题I只需要进行异或操作就可以完成，但这个题显然没有这么简单。我参考了Discuss点击打开链接，里面的分析还是非常有用的我贴出主要的
In Single Number, it is easy to think of XOR solution because XOR manipulation has such properties:

Commutative: A^B == B^A, this means XOR applies to unsorted arrays just like sorted. (1^2^1^2==1^1^2^2)

Circular: A^B^...^B == A where the count of B's is a multiple of 2.

So, if we apply XOR to a preceding zero and then an array of numbers, every number that appears twice will have no effect on the final result. Suppose there is a number H which appears just once, the final XOR result will be 0^H^...H where H appears as many
as in input array.

When it comes to Single Number II (every one occurs K=3 times except one occurs M times, where M is not a multiple of K), we need a substitute of XOR (notated as @) which satisfies:

Commutative: A@B == B@A.

Circular: A@B@...@B == A where the count of B's is a multiple of K.

We need to MAKE the @ operation. This general solution suggests that we maintain a state for each bit, where the state corresponds to how many '1's have appeared on that bit, so we need a int[32] array.
这可以从本质上解决这类题，但是链接里的第一个回答，我还是没有弄明白

3.Java AC代码

public int singleNumber(int[] nums) {
int k = 3;
int res = 0;
int[] bitCount = new int[32];
for (int i = 0; i < bitCount.length; i++) {
bitCount[i] = 0;
}
for (int i = 0; i < nums.length; i++) {
int tmp = nums[i];
for (int j = 0; j < bitCount.length; j++) {
int hasBit = tmp & (1 << j);
if (hasBit != 0) {
bitCount[j] = (bitCount[j] + 1) % k;
}

}
}
for (int i = 0; i < bitCount.length; i++) {
if (bitCount[i] > 0) {
res |= (1 << i);
}
}
return res;
}