Single number

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?

用个现成的 unordered_map, 算是hash table, O(N) buildup O(1) look up

class Solution {
public:
    int singleNumber(int A[], int n) {
        unordered_map<int,int> myMap;
        for (int i=0; i<n; i++)
        {
            if (myMap.count(A[i])==0)
                myMap[A[i]]=i;
            else
                myMap.erase(A[i]);
        }
        return myMap.begin()->first;
    }
};

如果不用额外buffer,放狗一搜，都是位操作。。。那玩意从来没明白过。有人说sort,不过sort 是NlgN. 没好哪里去。。。不过LEETCODE OJ能过。。。这也是奇葩的地方，呵呵。

class Solution {
public:
    int singleNumber(int A[], int n) {
        sort(A,A+n);
        bool found=false;
        int ind;
        for (int i=0; i<=n/2;i++)
        {
            if (A[2*i]!=A[2*i+1])
            {
                found=true;
                ind=2*i;
                break;
            }
        }
        if (found)
            return A[ind];
        else
            return A[n-1];
    }
};