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

• 题目：
  

  ---

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：
  

  ---

              1）已排序数组查找，二分查找

• 实现：
  

  ---

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>{-,-};

     }
 };

     2.  常规实现

 class Solution {
 public:
     vector<int> searchRange(int A[], int n, int target) {
         int low=,high=n-,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)/;
             if(A[middle]==target)
             {
                 isFind=true;
                 break;
             }
             else if(A[middle]<target)
                 low=middle+;
             else
                 high=middle-;
         }
         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>=&&A[low]==target) low--;//下界要>=0
             low++;
             while(high<=n-&&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(-);
             vec.push_back(-);
         }
         return vec;
     }
 };