leetcode题解:Search for a Range (已排序数组范围查找)

• 题目：

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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].

• 说明：

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              1）已排序数组查找，二分查找

• 实现：

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1. STL实现

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
       auto low_bound=lower_bound(A,A+n,target);//第一个大于等于>=target元素的指针位置
       auto up_bound=upper_bound(low_bound,A+n,target);//第一个大于>target元素的指针位置
       if(*low_bound==target)//target是否存在于A[]
       {
           return vector<int>{distance(A,low_bound),distance(A,prev(up_bound))};
       }
       else return vector<int>{-1,-1};
        
    }
};

     2.  常规实现

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        int low=0,high=n-1,middle;
        bool isFind=false;
        vector<int> vec;
        while(low<=high)//二分查找，直至找到，并置标志true
        {
            middle=(low+high)/2;
            if(A[middle]==target) 
            {
                isFind=true;
                break;
            }
            else if(A[middle]<target)
                low=middle+1;
            else
                high=middle-1;
        }
        if(isFind)//如果找到，确定开始、结束与target相等的元素位置
        {
            low=middle;
            high=middle;
            while(low>=0&&A[low]==target) low--;//下界要>=0
            low++;
            while(high<=n-1&&A[high]==target) high++;//上界要<=n-1
            high--;
            vec.push_back(low);
            vec.push_back(high);
        }
        else//没有目标值
        {
            vec.push_back(-1);
            vec.push_back(-1);
        }
        return vec;
    }
};