[leetcode] Search for a Range

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].

思路：二分查找

代码：

class Solution {public:    vector<int> searchRange(int A[], int n, int target) {        int start=0,end=n-1,mid=(start+end)/2,flag=0;        while(start<end){            if(A[mid]>target){end=mid-1;mid=(start+end)/2;}            if(A[mid]<target){start=mid+1;mid=(start+end)/2;}            if(A[mid]==target){flag=1;break;}        }        vector<int> res;        if(n==0){            res.push_back(-1);            res.push_back(-1);            return res;        }        if(n==1){            if(target==A[0]){                res.push_back(0);                res.push_back(0);                return res;            }else{                res.push_back(-1);                res.push_back(-1);                return res;            }        }        if(flag==0){            res.push_back(-1);            res.push_back(-1);            return res;        }         else{            int start=mid,end=mid;            while(A[start]==target){                start--;                if(start<0){                    start++;                    break;                }            }            while(A[end]==target){                end++;                if(end==n){                    end--;                    break;                }            }            if(start<0) start++;            if(end==n) end--;            if(A[start]!=target) start++;            if(A[end]!=target) end--;            res.push_back(start);            res.push_back(end);            return res;        }    }};