LeetCode-Single NumberI II III

I)

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

第一个比较好求解，异或操作就可以了：

    public int singleNumber(int[] nums) {        for (int i = 1; i < nums.length; i++) {nums[0] ^= nums[i];}return nums[0];    }

III)

在看第三个，第三个跟第一个的关系比较密切：

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:

1. The order of the result is not important. So in the above example, [5, 3] is also correct.
2. Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?

这个可以跟第一个联想，唯一不同的就是这个有两个数字出现一次。所以现在的目标就是转化到第一个，将数组分成两组，并且将两个数字分开，怎么分呢？

    public int[] singleNumber(int[] nums) {        int n = 0;        for (int i = 0; i < nums.length; i++) n ^= nums[i];        n &= ~(n-1);        int[] ret = new int[2];                for (int i = 0; i < nums.length; i++) {            if ((nums[i] & n) == 0) ret[0] ^= nums[i];            else ret[1] ^= nums[i];        }                   return ret;    }

代码如上，先求整个数组的异或，则最终等价于两个出现一次的数字异或，然后我们根据异或的结果进行分组，很显然。如果异或的二进制表示中，有一位为1，则表示。这两个数字在此位置的二进制表示是不同的。所以就可以根据这个位来进行分组，然后就把问题转化为第一个问题了！

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?

第二个，跟上面的关系不大，尽管都是位操作：

    public int singleNumber(int[] nums) {    int cntOne = 0;    int number = 0;    for (int i = 0; i < 32; i++) {    cntOne = 0;    for (int j = 0; j < nums.length; j++) {    if ((nums[j] & (1 << i)) != 0) cntOne++;    }    number |= (cntOne %= 3) << i;    }    return number;    }

对int的每一位进行遍历判断。此时遍历数组统计每个位1,出现的次数，然后对3取余数就是出现一次的那个数在本位的次数，然后将这个32个0或1组合起来！