lintcode:Search for a Range

Given a sorted array of n integers, find the starting and ending
position of a given target value.

If the target is not found in the array, return [-1, -1].

Example Given [5,
7, 7, 8, 8, 10] and target value 8, return [3, 4].

Challenge O(log n) time.

关于上界和下界详细可以见我的这篇博文：​​二分查找上界和下界​​

class Solution {
    /**
    *@param A : an integer sorted array
    *@param target :  an integer to be inserted
    *return : a list of length 2, [index1, index2]
    */
public:
    vector<int> searchRange(vector<int> &A, int target) {
        // write your code here

        int len = A.size();
        vector<int> res(2);

        if (len == 0){
            res[0] = res[1] = -1;
            return res;
        }

     int l = 0, r = len;

    while (l < r){
        int mid = l + (r - l) / 2;
        if (target == A[mid]){
            r = mid;
        }
        else if (target>A[mid]){
            l = mid + 1;
        }
        else{
            r = mid;
        }
    }

    if(A[l]==target){
        res[0]=l;
    }else{
        res[0]=-1;
    }


    l = 0, r = len;

    while (l<r){
        int mid = l + (r - l) / 2;
        if (target == A[mid]){
            l = mid+1;
        }
        else if (target>A[mid]){
            l = mid + 1;
        }
        else{
            r = mid;
        }
    }

    if(A[r-1]==target){
        res[1]=r-1;
    }else{
        res[1]=-1;
    }


        return res;
    }
};