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：

1.其实就是 if target value duplicated， find the duplicated range

二分找到target，左右遍历找index上下界；没找到，就返回[-1,-1]

 2. 在标准binary search修改，先找left one；然后找right one。

比较: 方法1的最坏情况是，O(n) :[1,1,1,1,1,1] find 1; 一般情况下是O(lgn+r)

方法2的最坏情况是，target only appear once；最坏最好都是 O(2lgn);可以增加判断，将only once的情况改进成O(lgn);

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target) {
        int low=0;
        int hi=n-1;
        int mid=0;
        int index;
        vector<int> range;
        while(low<=hi){
            mid=(low+hi)/2;
            if(target==A[mid]){ index=mid;break;}
            if(target<A[mid]) hi=mid-1;
            else low=mid+1;
        }
        if(hi<low) {
            range.push_back(-1);
            range.push_back(-1);
            return range;
        }
        
        int indexLow=index;
        int indexHi=index;
        while(indexHi<n-1){
            if(A[indexHi]==A[indexHi+1]) indexHi++;
            else break;
        }
        while(indexLow>0){
            if(A[indexLow]==A[indexLow-1]) indexLow--;
            else break;
        }
        range.push_back(indexLow);
        range.push_back(indexHi);
        return range;
    }
};