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].

给一个排序的数组，找出目标值的范围，若目标值不存在，返回[-1,-1].

代码如下：

class Solution {
public:
    vector<int> searchRange(int A[], int n, int target)
    {
        int begin = 0;
        int end = n-1;
        vector<int> vs;
        while (begin <= end)
        {
            int mid = (begin + end) / 2;
            if (A[mid] == target)
            {
                int min = mid;
                while (min >= 0 && A[mid] == A[min])
                    min--;
                int max = mid;
                while (max <= end && A[mid] == A[max])
                {
                    max++;
                }
                vs.push_back(min+1);
                vs.push_back(max - 1);
                return vs;
            }
            else if (A[mid] > target)
            {
                end = mid - 1;
            }
            else
            {
                begin = mid + 1;
            }
        }
        vs.push_back(-1);
        vs.push_back(-1);
        return vs;
    }
};

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.
You may assume no duplicates in the array.

Here are few examples.
[1,3,5,6], 5 → 2
[1,3,5,6], 2 → 1
[1,3,5,6], 7 → 4
[1,3,5,6], 0 → 0

若当前数字大于或等于目标值，则返回当前坐标，如果遍历结束了，说明目标值比数组中任何一个数都要大，则返回数组长度n即可
代码：

class Solution {
public:
    int searchInsert(int A[], int n, int target)
    {
        if (A[n-1]< target)
            return n;
        int left = 0, right = n - 1;
        while (left < right)
        {
            int mid = left + (right - left) / 2;
            if (A[mid] == target)
                return mid;
            else if (A[mid] < target)
                left = mid + 1;
            else
                right = mid;
        }
        return right;
    }
};