[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].

问题描述：给定一个排序的整数数组，找到一个给定值的起始和终止位置，算法的时间复杂度是O(log n)。

由于算法的时间复杂度是O(log n)，因此，可以采用二分查找，只不过当查找到给定的值时，要向左和向右进行延伸，看是否有重复的元素。

class Solution {public:    vector<int> searchRange(int A[], int n, int target) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        int left = 0, right = n - 1, mid = 0;        vector<int> vec;                while(left <= right) {            mid = (left + right) / 2;            if(target > A[mid])                left = mid + 1;            else if(target < A[mid])                right = mid - 1;            else {                int i = mid;                while(i >= left && A[i] == target)                    --i;                                int j = mid;                while(j <= right && A[j] == target)                    ++j;                                vec.push_back(++i);                vec.push_back(--j);                                return vec;            }        }                vec.push_back(-1);        vec.push_back(-1);                return vec;    }};