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].

二分查找的变形：

1、没有找到返回[-1, -1]

2、只存在1个返回[pos, pos]

3、存在多个，返回端点[leftPos, rightPos]

#include <iostream>
#include <set>
#include <map>
#include <vector>
#include <sstream>
#include <string>
#include <algorithm>
#include <bitset>
using namespace std;

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int>vec;
        int leftPos = BinarySearchLeft(nums, target);
        int rightPos = BinarySearchRight(nums, target);
        if (nums[leftPos] != target)
        {
            vec.push_back(-1);
            vec.push_back(-1);
        }
        else
        {
            vec.push_back(leftPos);
            vec.push_back(rightPos);
        }
        return vec;
    }
private:
	int BinarySearchLeft(vector<int>& nums, int target)
    {
        int left = 0;
        int right = nums.size() - 1;
        while(left <= right)
        {
            int mid = left + ((right - left) >> 1);
          
            if (nums[mid] >= target)
            {
                right = mid - 1;
            }
            else
            {
                left = mid + 1;
            }
        }
        return left;
    }

    int BinarySearchRight(vector<int>& nums, int target)
    {
        int left = 0;
        int right = nums.size() - 1;
        while(left <= right)
        {
            int mid = left + ((right - left) >> 1);
          
            if (nums[mid] > target)
            {
                right = mid - 1;
            }
            else
            {
                left = mid + 1;
            }
        }
        return right;
    }
};

int main()
{
	Solution s;
	vector<int>vec{0,0,0,1,2,3};
	vector<int>v = s.searchRange(vec, 0);
	for (auto it = v.begin(); it != v.end(); it++)
	{
		cout << *it << endl;
	}
	return 0;
}

看，这个时间，估计得有更高效的算法了。