Single Element in a Sorted Array

题目540.

Single Element in a Sorted Array

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• Total Accepted: 4418
• Total Submissions: 8253
• Difficulty: Medium
• Contributors:rajaditya

Given a sorted array consisting of only integers where every element appears twice except for one element which appears once. Find this single element that appears only once.

解题思路：从数组递增有序和O(log n)时间复杂度推断，题目可以采用二分查找求解。

初始令左、右指针lo, hi分别指向0, len(nums) - 1当lo < hi时执行循环：令mi = (lo + hi) / 2若nums[mi] == nums[mi - 1]：数组可以分为[lo, mi - 2], [mi + 1, hi]两部分，目标元素位于长度为奇数的子数组中。同理，若nums[mi] == nums[mi + 1]：数组可以分为[lo, mi - 1], [mi + 2, hi]两部分，目标元素位于长度为奇数的子数组中。若nums[mi]与nums[mi - 1], nums[mi + 1]均不相等，则返回nums[mi]

代码：

#include <iostream>
#include <vector>
using namespace std;
int singleNonDuplicate(vector<int>& nums)
{
    if(nums.size() == 1)
        return nums[0];
    if(nums[0] != nums[1])
        return nums[0];
    if(nums[nums.size()-1] != nums[nums.size()-2])
        return nums[nums.size()-1];
    int l = 0, r = nums.size()-1;
while(l <= r)//"=" is must......
    {
        int mid = (l+r)/2;
        if(nums[mid]!=nums[mid-1] && nums[mid]!=nums[mid+1])
            return nums[mid];
        if(nums[mid] == nums[mid+1])//change the array to two parts:[l,mid-1]&&[mid+2,r];
        {
            if((mid-l) % 2 == 1)
                r = mid-1;
            else
                l = mid+2;
        }
        else if(nums[mid] == nums[mid-1])
        {
            if((mid-l-1) % 2 == 1)
                r = mid-2;
            else
                l = mid+1;
        }
    }
    /*
    if(nums[0] != nums[1])
    {
        return nums[0];
    }
    else if(nums[nums.size()-1] != nums[nums.size()-2])
    {
        return nums[nums.size()-1];
    }
    else
    {
        for(int i=1;i<nums.size();i++)
        {
            if(nums[i]!=nums[i-1] && nums[i]!=nums[i+1])
            {
                return nums[i];
            }
        }
    }*/
    return -1;
}


int main()
{
    int t;
    vector<int> v;
    while(cin>>t)
        v.push_back(t);
    cout<<singleNonDuplicate(v)<<endl;


    return 0;
}