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].
class Solution { public: vectorsearchRange(int A[], int n, int target) { vector ivec; int left=lower_bound(A,n,target); if(left==n||A[left]!=target) { ivec.push_back(-1); ivec.push_back(-1); return ivec; } ivec.push_back(left); int right=upper_bound(A,n,target); ivec.push_back(right-1); return ivec; } //可插入target的第一個合適位置 int lower_bound(int A[],int n,int target){ int first=0,len=n,half; int mid; while(len>0){ half=len>>1; mid=first+half; if(A[mid] 0){ half=len>>1; mid=first+half; if(A[mid]>target){ len=half; } else{ first=mid+1; len=len-half-1; } } return first; } };