The problem:

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].

My analysis:

The idea behind this question is very elegant and tricky.
key: when the target value is found, we are not return it (very important). we keep going to reach the the boundry.
It‘s very skllful to write the right checking condition, and read the right value from the proper pointer.
1. When we reach the the left most target, we stop the low pointer(left pointer).
if (A[mid] < target) //we move the low pointer only when A[mid] is strictly smaller than target.
low = mid + 1;
else //A[mid] <= target, we keep move high pinter, even when A[mid] == target.
high = mid - 1

2. The same idea could be used to find the right most target, we stop the high pointer(right pointer).
if (A[mid] <= target)
low = mid + 1;
else //A[mid] > target, we move high pointer only when A[mid] is strictly larger than target.
high = mid - 1;

My solution:

public class Solution {    public int[] searchRange(int[] A, int target) {                int[] ret = new int[2];        ret[0] = -1;        ret[1] = -1;                if (A.length == 0)            return ret;                int llow = 0;        int lhigh = A.length - 1;        int rlow = 0;        int rhigh = A.length - 1;        int mid = -1;                while (llow <= lhigh) {            mid = (llow + lhigh) / 2;                        if (A[mid] < target) //the idea behind this approaching is very tricky and elegant!                llow = mid + 1;            else                 lhigh = mid - 1;        }                while (rlow <= rhigh) {            mid = (rlow + rhigh) / 2;                        if (A[mid] <= target)                rlow = mid + 1;            else                 rhigh = mid - 1;        }                if (llow <= rhigh) {            ret[0] = llow;            ret[1] = rhigh;        }                return ret;     }}

[LeetCode#34]Search for a Range