[LeetCode]136. Single Number&137. Single Number II&260. Single Number III

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本文介绍了三种寻找数组中唯一出现一次的元素的算法：SingleNumber、SingleNumberII 和 SingleNumberIII。这些算法分别针对元素出现两次、三次和其他元素出现两次的情况，并且均实现了线性时间复杂度的要求。

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136 . Single Number
Easy

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?

1ms:

 public class Solution {
    public int singleNumber(int[] nums) {
          int res = 0;
         for(int a:nums){
             res ^=a;
         }
         return res;
    }
}

137 . Single Number II
Medium

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?

1ms:

public int singleNumber(int[] nums) {
        int ones = 0, twos = 0;
        for(int i = 0; i < nums.length; i++){
            ones = (ones ^ nums[i]) & ~twos;
            twos = (twos ^ nums[i]) & ~ones;
        }
        return ones;
    }

260 . Single Number III
Medium

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

1ms:

public int[] singleNumber(int[] nums) {
        int[] re = {0,0};
        int c = 0;
        for(int i=0;i<nums.length;i++){
            c ^= nums[i];
        }
        int p = 1;
        while(true){
            if((p&c)!=0)
                break;
            else
                p = p<<1;
        }
        for(int i=0;i<nums.length;i++){
            if((p&nums[i])==0)
                re[0] ^= nums[i];
            else
                re[1] ^= nums[i];
        }
        return re;
    }