题目：
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].
思路：
二分查找
package array; public class SearchForARange { public int[] searchRange(int[] nums, int target) { int[] res = { -1, -1 }; int n; if (nums == null || (n = nums.length) == 0) return res; int index = -1; int start = 0; int end = n - 1; while (start <= end) { int mid = (end - start) / 2 + start; if (nums[mid] == target) { index = mid; break; } else if (nums[mid] < target) { start = mid + 1; } else { end = mid - 1; } } res[0] = index; res[1] = index; while (res[0] > 0 && nums[res[0] - 1] == target) --res[0]; while (res[1] < n - 1 && nums[res[1] + 1] == target) ++res[1]; return res; } public static void main(String[] args) { // TODO Auto-generated method stub int[] nums = { 5, 7, 7, 8, 8, 10 }; SearchForARange s = new SearchForARange(); for(int i : s.searchRange(nums, 10)) System.out.println(i); } }
原文：http://www.cnblogs.com/shuaiwhu/p/5068724.html