Leetcode--Search for a Range

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

Your algorithm ' s runtime complexity must is in the order of O(log n).

if The target is not a found in the array, Return [-1,-1] .

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

Analysis: because there are sequential sequences, it should be possible to find by means of dichotomy. Two recursive exits: (1) n=1 and no targets found; (2) The target has been found.
Note the case of TARGET>A[N/2], when the [A+N/2, N-N/2] sub-sequence is re-recursive lookup, the return value of the target position to add the previous N/2.

Class Solution {public:    vector<int> searchrange (int a[], int n, int target) {        vector<int> ret;        if (n = = 1 && target!=a[0])   //Recursive exit 1, the target value            ret.assign (2,-1) is not found;        else if (target = = A[n/2]) {      //Recursive exit 2, locate the target value            int Tbeg = N/2, tend = N/2;            while ((--tbeg) >= 0 && target = = A[tbeg]);            while ((++tend) < n && target = = A[tend]);            Ret.push_back (Tbeg + 1);            Ret.push_back (tend-1);        }        else if (target < A[N/2])            ret = Searchrange (A, N/2, target);        else{    ret = searchrange (A + N/2, N-N/2, target);    if (ret[0] = =-1 && ret[1] = = 1)  return ret;    Ret[0] + = N/2; RET[1] + = N/2;       attention***        }        return ret;}    ;

Leetcode--Search for a Range