LeetCode--Search for a Range

最新推荐文章于 2025-09-11 19:19:55 发布

Leon__CAI

最新推荐文章于 2025-09-11 19:19:55 发布

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分类专栏： leetcode 文章标签： leetcode search binary search

本文链接：https://blog.youkuaiyun.com/shaya118/article/details/42611945

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本文介绍了一种算法，用于在一个已排序的整数数组中查找给定目标值的起始和结束位置。该算法的时间复杂度为O(log n)，并提供了一个具体的C++实现示例。

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(int A[], int n, int target) 
    {
        vector<int> res;
        if(n == 0 )
            return res;
        int low = search_low(A,0,n-1,target);
        res.push_back(low);
        if(low == -1)
            res.push_back(low);
        else
        {
            int high = search_high(A,0,n-1,target);
            res.push_back(high);
        }
        return res;
    }
    int search_low(int* a, int low, int high, int target)
    {
        int loc = -1;
        while(low <= high)
        {
            int mid = (low+high)/2;
            if(a[mid] == target)
            {
                loc = mid;
                high = mid-1;
            }
            else if(a[mid] > target)
                high = mid-1;
            else
                low = mid+1;
        }
        return loc;
    }
    int search_high(int* a, int low, int high, int target)
    {
        int loc = -1;
        while(low <= high)
        {
            int mid = (low+high)/2;
            if(a[mid] == target)
            {
                loc = mid;
                low = mid+1;
            }
            else if(a[mid] > target)
                high = mid-1;
            else
                low = mid+1;
        }
        return loc;
    }
};