【LeetCode 34】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].

题意：

　　给定一个有序数组，其中可能包含一些重复数字，给定一个数字，求其第一次和最后一次出现的坐标，要求 O(log n)。

思路：

　　有序数组+ O(log n)，很显然二分搜索啦。

C++：

class Solution {
public:

    int Bsearch(vector<int>& nums, int len, int start, int end, int tar, bool flag)
    {
        if(start > end)
            return -1;
        
        int mid = (start + end) / 2;
        
        if(nums[mid] == tar)
        {
            if(flag)
            {
                if((mid > 0 && nums[mid - 1] != tar) || mid == 0)
                {
                    return mid;
                }
                else
                {
                    return Bsearch(nums, len, start, mid - 1, tar, flag);
                }
            }
            else
            {
                if((mid < len - 1 && nums[mid + 1] != tar) || mid == len - 1)
                {
                    return mid;
                }
                else
                {
                    return Bsearch(nums, len, mid + 1, end, tar, flag);
                }
            }

        }
        else if(nums[mid] < tar)
        {
            return Bsearch(nums, len, mid + 1, end, tar, flag);
        }
        else
        {
            return Bsearch(nums, len, start, mid - 1, tar, flag);
        }
    }

    vector<int> searchRange(vector<int>& nums, int target) {
        
        int len = nums.size();
        vector<int> ret;
        if(len == 0)
            return ret;
        
        ret.push_back(Bsearch(nums, len, 0, len - 1, target, true));
        ret.push_back(Bsearch(nums, len, 0, len - 1, target, false));
        
        return ret;
    }
};