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

 1 class Solution {
 2 public:
 3     vector<int> searchRange(int A[], int n, int target) {
 4        auto low_bound=lower_bound(A,A+n,target);//第一个大于等于>=target元素的指针位置
 5        auto up_bound=upper_bound(low_bound,A+n,target);//第一个大于>target元素的指针位置
 6        if(*low_bound==target)//target是否存在于A[]
 7        {
 8            return vector<int>{distance(A,low_bound),distance(A,prev(up_bound))};
 9        }
10        else return vector<int>{-1,-1};
11
12     }
13 };

     2.  常规实现

 1 class Solution {
 2 public:
 3     vector<int> searchRange(int A[], int n, int target) {
 4         int low=0,high=n-1,middle;
 5         bool isFind=false;
 6         vector<int> vec;
 7         while(low<=high)//二分查找，直至找到，并置标志true
 8         {
 9             middle=(low+high)/2;
10             if(A[middle]==target)
11             {
12                 isFind=true;
13                 break;
14             }
15             else if(A[middle]<target)
16                 low=middle+1;
17             else
18                 high=middle-1;
19         }
20         if(isFind)//如果找到，确定开始、结束与target相等的元素位置
21         {
22             low=middle;
23             high=middle;
24             while(low>=0&&A[low]==target) low--;//下界要>=0
25             low++;
26             while(high<=n-1&&A[high]==target) high++;//上界要<=n-1
27             high--;
28             vec.push_back(low);
29             vec.push_back(high);
30         }
31         else//没有目标值
32         {
33             vec.push_back(-1);
34             vec.push_back(-1);
35         }
36         return vec;
37     }
38 };

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