136. Single Number

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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?

Solution 1:

思路： 用set检验是不是有重复的。把没有重复的加到set里面，有重复的从set里面remove掉 .最后set里面只存在一个数，用iterator给读出来即可。

注意corner case: 数组只有一个数的时候就直接读出。

Time Complexity: O(N)

 public int singleNumber(int[] nums) {

        if(nums.length==1)

        {

            return nums[0];

        }

        int number=0;

        Set<Integer> res=new HashSet<Integer>();

        for(int i=0;i<nums.length;i++)

        {

            if(!res.contains(nums[i]))

            {

                res.add(nums[i]);

            }

            else

            {

                res.remove(nums[i]);

                

            }

        }

        for(int single: res)

        {

            number=single;

        }

        return number;

 

    }

Solution 2:

没有想出来不用extra space的，参考了discussion

https://discuss.leetcode.com/topic/47560/xor-java-solution/5

要熟悉XOR的工作原理

Reference: http://www.programmerinterview.com/index.php/java-questions/xor-in-java/

X     Y     Output

0     0     0

0     1     1

1     0     1

1     1     0 

Tricks: 0^a=a 2,a^b=b^a 3,a^a=0, a^1=a(inversion)

题目思路：because the xor has three properties. 1,0^a=a 2,a^b=b^a 3,a^a=0
so swapping the number would not change the output at the end, we can see the list as all the same numbers are adjacent. the same numbers would be 0 after xor. the one remaining would be the answer.

因为除了一个数之外其他的数都有一个相同的数，所以最后剩下来的就是 0^a=a极其巧妙.

    public int singleNumber(int[] nums) {

        int count=0;

        for(int i=0;i<nums.length;i++)

        {

            count^=nums[i];

        }

        return count;

}

136. Single Number

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原文地址：http://www.cnblogs.com/Machelsky/p/5856283.html