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].
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因为原数组是有序的，二分查找后，前后遍历即可。
1 public class Solution { 2 public int[] searchRange(int[] nums, int target) { 3 int index = Arrays.binarySearch(nums,target); 4 if(index < 0) return new int[]{-1,-1}; 5 int s = index - 1,e = index + 1; 6 while(s >= 0 && nums[s] == nums[index]) s--; 7 while(e < nums.length && nums[e] == nums[index]) e++; 8 return new int[]{++s,--e}; 9 } 10 }