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最新推荐文章于 2019-11-02 19:59:35 发布

原创最新推荐文章于 2019-11-02 19:59:35 发布 · 459 阅读

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本文介绍了一种使用二分法在已排序整数数组中查找指定目标值起始和结束位置的算法。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。

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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].

二分法的应用

 vector<int> searchRange(vector<int>& nums, int target) {
        int n=nums.size();
        int left=0,right=n-1,mid;
        vector<int> res;
        
        while(left<=right)    //寻找左边界，即nums[mid]=target并且nums[mid-1]不等于target
        {
            mid=(left+right)/2;
            if(nums[mid]<target)
                left=mid+1;
            else if(nums[mid]>target)
                right=mid-1;
            else 
            {
                if(mid==0||nums[mid-1]!=target)
                {
                    res.push_back(mid);
                    break;
                }
                else right=mid-1;
            }
        }
        
        if(left>right)
         {
            res.push_back(-1);
            res.push_back(-1);
             return res;
         }
        left=0;right=n-1;
        while(left<=right)//寻找右边界，即nums[mid]=target并且nums[mid+1]不等于target
        {
            mid=(left+right)/2;
            if(nums[mid]<target)
                left=mid+1;
            else if(nums[mid]>target)
                right=mid-1;
            else 
            {
                if(mid==n-1||nums[mid+1]!=target)
                 {
                     res.push_back(mid);
                     break;
                 }
                 else left=mid+1;
            }
        }  
        return res;
        
    }