LeetCode 034. Search for a Range

Search for a Range

 

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].


這道題目要求O(log n)的複雜度,那肯定用二分查找。一開始想到的是先二分查找,然後再左右擴展。代碼如下:

class Solution 
{
public:
    vector<int> searchRange(int A[], int n, int target) 
    {
        vector<int> result;
        if(n<=0)
        {
            result.push_back(-1);
            result.push_back(-1);
            return result;
        }
        int low = 0;
        int high = n-1;
        int mid;
        while(low <= high)
        {
            mid = (low+high)/2;
            if(A[mid]==target)
                break;
            else if(A[mid] >= target)
                high = mid - 1;
            else
                low = mid + 1;
        }
        
        if(A[mid] != target)
        {
            result.push_back(-1);
            result.push_back(-1);
            return result;           
        }
        else
        {
            low = high = mid;
            while(low >= 1)
            {
                low -- ;
                if(A[low]!=target)
                {
                    low ++;
                    break;
                }
            }
            while(high <= n-2)
            {
                high ++ ;
                if(A[high]!=target)
                {
                    high --;
                    break;
                }
            }
            result.push_back(low);
            result.push_back(high);
            return result;      
            
        }
    }
};

這樣Accept沒問題,時間是74ms,但是這個擴展複雜度很難講,萬一一個數組全是target,那麼擴展複雜度是O(n),這肯定是不合適的。

爲此呢,我想到的是以第一查找到的下標爲軸,然後將軸的兩邊分別查找target,直到返回結果是-1或者到達邊界才退出。

代碼如下:

class Solution 
{
public:
    int BinarySearch(int A[],int low, int high, int target)
    {
        int mid;
        while(low<=high)
        {
            mid = (low+high)/2;
            if(A[mid]==target)
                return mid;
            else if(A[mid]>target)
                high = mid-1;
            else
                low = mid+1;
        }
        return -1;
    }
    
    vector<int> searchRange(int A[], int n, int target) 
    {
        vector<int> result(2,-1);
        if(n<=0)
            return result;
        int pos;
        int temp;
        int start,end;
        pos = BinarySearch(A, 0, n-1, target);
        if(pos == -1)
            return result;
        else
            start = end = pos;
            
        temp = pos;
        while(temp >= 0)
        {
            temp = BinarySearch(A,0,temp-1,target);
            if(temp == -1)
                break;
            else
                start = temp;
        }
        
        temp = pos;
        while(temp < n)
        {
            temp = BinarySearch(A,temp+1, n-1, target);
            if(temp == -1)
                break;
            else
                end = temp;
        }
        result[0] = start;
        result[1] = end;
        return result;
    }
};

時間是52ms.




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