Leetcode: Single Number II

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?

思路和Single Number 类似

思路1：先排序，后寻找是否三个连续值相同，不同则return这个值。用到Arrays.sort函数，时间复杂度不符合要求。

Solution1:

public class Solution {
public int singleNumber(int[] A) {
if (A == null || A.length == 0) {
return -1;
}
if (A.length < 3) {
return A[0];
}

Arrays.sort(A);
for (int i = 0; i < A.length; i++) {
if (i + 2 >= A.length - 1 || A[i] != A[i + 2]) {
return A[i];
}
i += 2;
}

return -1;
}
}

思路2：使用三进制异或。存三个变量，ones存出现过一次的数，twos存出现过两次的数，threes则判断这个数是否同时在ones和twos里面出现，若是则说明该数出现了三次，这时需要将这个数从ones和twos中删除。最后的出来的ones就是结果。

Solution2:

public class Solution {
public int singleNumber(int[] A) {
if (A == null || A.length == 0) {
return -1;
}

int ones = 0, twos = 0, threes = 0;
for (int i = 0; i < A.length; i++) {
twos |= ones & A[i];
ones ^= A[i];
threes = ones & twos;
ones &= ~threes;
twos &= ~threes;
}

return ones;
}
}

 思路3：模拟三进制异或。即：各个数的某个二进制位之和能被3整除，则说明要寻找的single number在这一位上位0，反之为1。

Solution3：

public int singleNumber(int[] A) {
if (A == null || A.length == 0) {
return -1;
}

int res = 0;
int[] bit = new int[32];
for (int i = 0; i < 32; i++) {
for (int j = 0; j < A.length; j++) {
bit[i] += A[j] >> i & 1;
bit[i] %= 3;
}
res |= bit[i] << i;
}

return res;
}