34. Search for a Range

最新推荐文章于 2022-05-19 15:18:54 发布

tanlichun789

最新推荐文章于 2022-05-19 15:18:54 发布

阅读量146

点赞数

分类专栏： LeetCode

LeetCode 专栏收录该内容

59 篇文章

订阅专栏

本文介绍了一种算法，用于在一个升序排列的整数数组中找到给定目标值的起始和结束位置。该算法遵循O(log n)的时间复杂度要求，并提供了一个C++实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given an array of integers sorted in ascending order, 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].

如下写法太丑陋了，意思是先二分找到第一个K 然后左右再顺序查找扩展，如果遇到数组都是 KKKKKK 那么时间复杂度还是0(N) ,如下写法加快了些，并不好，直到我在剑指offer 看到了类似的题目：

#include <iostream> 
#include <vector>

using namespace std;

class Solution {
public:
	vector<int> searchRange(vector<int>& nums, int target) {
		vector<int> volidAns = { -1, -1 };
		if (!nums.size())  return volidAns;
		vector<int> ans;
		int min = 0;
		int max = nums.size() - 1;
		int mid;
		while (min <= max)
		{
			mid = (min + max) / 2;
			if (nums[mid] == target)
			{
				search(mid, ans, nums);
				return ans;
			}
			if (nums[mid] < target) min = mid + 1;
			else
			{
				max = mid - 1;
			}

		}
	}
private:
	void search(int index, vector <int> & ans, vector<int>& nums)
	{
		int rightIndex = index;
		int right = 1, left = 1;
		int leftIndex = index;
		int target = nums[index];
		while (leftIndex >= 0 && nums[leftIndex] == target)
		{
			leftIndex -= left;
			left *= 2;
		}
		left /=2;
		leftIndex += left;
		while (leftIndex >= 0 && nums[leftIndex] == target) --leftIndex;
		++leftIndex;
		ans.push_back(leftIndex);

		while (rightIndex < nums.size() && nums[rightIndex] == target)
		{
			rightIndex += right;
			right *=2;
		}
		right /= 2;
		rightIndex -= right;
		while (rightIndex < nums.size() && nums[rightIndex] == target) ++rightIndex;
		--rightIndex;
		ans.push_back(rightIndex);

	}
};

int main()
{
	vector<int> haha = { 1 };
	Solution a; 
	vector <int> ans = a.searchRange(haha, 0);
	for (int i = 0; i < ans.size(); ++i)
	{
		cout << ans[i] << " ";
	}
	system("pause");
	return 0; 
}

offer 写的比较棒：