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 ofO(logn).

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 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

class Solution {
public:
    vector searchRange(vector& nums, int target) {
        vectorvec;
        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& 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& 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;
	vectorvec{0,0,0,1,2,3};
	vectorv = s.searchRange(vec, 0);
	for (auto it = v.begin(); it != v.end(); it++)
	{
		cout << *it << endl;
	}
	return 0;
}
34. Search for a Range_第1张图片

看,这个时间,估计得有更高效的算法了。

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