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].

 

这题比较简单，就是边界的时候范围需要注意一下，我的方法先用二分法找到那个目标所在的位置，然后从那个位置增加和减少index直到不等于目标，于是就得到一个索引范围。代码如下：

1 class Solution { 2 public: 3     vector
     
       searchRange(vector
      
       & nums, int target) { 4         int len = nums.size(); 5         int start = 0; 6         int end = len; 7         vector
       
         result; 8         findStartEnd(nums,start,end,target,result); 9         if(result.size()==0){10             result.push_back(-1);11             result.push_back(-1);12         }13         return result;14     }15     16     void findStartEnd(vector
        
         &nums,int start,int end,int target,vector
         
          & result){17          int mid = (start+end)/2;18          if(start<=mid&&end>=mid){19          if(nums[mid]>target){20              if(start == mid) mid = mid-2;21              findStartEnd(nums,start,mid,target,result);22          }23          else if(nums[mid]
          
           < end ;i++){30 if(nums[i]!=nums[mid]){31 tail = i-1;32 break;33 }34 tail = i;35 }36 for(int j = mid-1 ;j>=start;j--){37 if(nums[j]!=nums[mid]){38 front = j+1;39 break;40 }41 front = j;42 }43 result.push_back(front);44 result.push_back(tail);45 }46 }47 }48 };