leetcode- Search for a Range

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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(vector<int>& nums, int target) {
        int begin=-1, end=-1;
        int low=0, high=nums.size()-1, mid=0;
        while(low<=high)
        {
            mid = low+(high-low)/2;
            if(nums[mid] > target)
            {
                high = mid-1;
            }
            else if(nums[mid] < target)
            {
                low = mid+1;
            }
            else if(nums[mid] == target)
            {
                begin = mid;
                high = mid-1;
            }
        }
        
        low = begin;
        high = nums.size()-1;
        if(low!=-1)
        {
            while(low<=high)
            {
                mid = low+(high-low)/2;
                if(nums[mid] > target)
                {
                    high = mid-1;
                }
                else if(nums[mid] < target)
                {
                    low = mid+1;
                }
                else if(nums[mid] == target)
                {
                    end = mid;
                    low = mid+1;
                }
            }
        }
        vector<int> res;
        res.push_back(begin);
        res.push_back(end);
        return res;
    }
};

leetcode- Search for a Range

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原文地址：http://my.oschina.net/u/2368952/blog/527549