LeetCode 136. Single Number

原题链接在这里：https://leetcode.com/problems/single-number/

题目：

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?

题解：

首先会想到HashMap，HashSet的想法，但会用到extra O(n) space.

所以就要用到bit manipulation，这里和Single Number II非常相似, 32位的int, 每一位的count若是出现奇数，那么这一位就是用来组成single number的.

但这里有个更快的方法，就是用异或 ^ operator. 异或 every number in the array and the result would be the single number.

Method 1 Time Complexity: O(n). Space: O(n).

Method 2 Time Complexity: O(n). Space: O(1).

Method 3 Time Complexity: O(n). Space: O(1).

AC Java:

public class Solution {
    public int singleNumber(int[] nums) {
        /*Method 1
        if(nums == null || nums.length == 0)
            return Integer.MIN_VALUE;
        
        HashSet hs = new HashSet();
        for(int i = 0; i < nums.length; i++){
            if(!hs.contains(nums[i])){
                hs.add(nums[i]);
            }else{
                hs.remove(nums[i]);
            }
        }
        
        Iterator it = hs.iterator();
        int res = Integer.MIN_VALUE;
        while(it.hasNext()){
            res = (int)it.next();
        }
        
        return res;
        */
        
        /*Method 2
        if(nums == null || nums.length == 0)
            return Integer.MIN_VALUE;
            
        int [] bitCounter = new int[32];
        for(int i = 0; i<32; i++){
            for(int j = 0; j<nums.length; j++){
                bitCounter[i] += (nums[j]>>i&1); //error
            }
        }
        int res = 0;
        for(int i = 0; i<32; i++){
            res += (bitCounter[i]%2)<<i;
        }
        return res;
        */
        
        //Method 3
        if(nums == null || nums.length == 0)
            return Integer.MIN_VALUE;
        int res = 0;
        for(int j = 0; j < nums.length; j++){
            res = res ^ nums[j];
        }
        return res;
    }
}

跟上Single Number II, Single Number III.