题目描述：
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].
结题思路：
采用二分法查找，如果找到再从这个位置向两边扩散，直到到达目标值的两边边界。
代码如下：
public class Solution {
public int[] searchRange(int[] nums, int target) {
int[] result = { -1, -1 };
int left = 0;
int right = nums.length - 1;
int mid, low, high;
if (target > nums[right] || target < nums[left])
return result;
while (left <= right) {
mid = (left + right) / 2;
if (nums[mid] == target) {
low = mid;
high = mid;
while (low >= 0 && nums[low] == target)
low--;
result[0] = low + 1;
while (high < nums.length && nums[high] == target)
high++;
result[1] = high - 1;
return result;
} else if (nums[mid] > target) {
right = mid - 1;
} else {
left = mid + 1;
}
}
return result;
}
}
Java [leetcode 34]Search for a Range
原文：http://www.cnblogs.com/zihaowang/p/4960203.html