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].
大致题意:找到值为target的下标范围
vector<int> searchRange(int A[], int n, int target) { int l,r,mid; std::vector<int> ans; l=0; r=n-1; //找到第一个不小于target的位置 while (l<=r) { mid=(l+r)>>1; if (A[mid]>=target) r=mid-1; else l=mid+1; } if (A[l]==target) ans.push_back(l); else{ ans.push_back(-1); ans.push_back(-1); return ans; } l=0; r=n-1; //找到第一个大于target的位置 while (l<=r) { mid=(l+r)>>1; if (A[mid]>target) r=mid-1; else l=mid+1; } ans.push_back(l-1); return ans; }
原文：http://www.cnblogs.com/mintmy/p/4190597.html