LeetCode--Search for a Range

标签：leetcode   search   binary search   

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].

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

LeetCode--Search for a Range

标签：leetcode   search   binary search   

原文地址：http://blog.csdn.net/shaya118/article/details/42611945