136. 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?

链接： http://leetcode.com/problems/single-number/

题解：

位运算，用异或处理掉出现两次的数字即可。

Time Complexity - O(n)， Space Complexity - O(1)。

public class Solution {
    public int singleNumber(int[] A) {
        int result = 0;
        
        for(int i : A){
            result ^= i;
        }
        
        return result;
    }
}

Update:

public class Solution {
    public int singleNumber(int[] nums) {
        if(nums == null || nums.length == 0)
            return 0;
        int res = 0;
        
        for(int i : nums)
            res ^= i;
            
        return res;
    }
}

二刷：

使用异或。a ^ b ^ a = b,  0 ^ b = b

Java:

Time Complexity - O(n)， Space Complexity - O(1)。

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

三刷:

Java:

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

 

测试：