34. Search for a Range java solutions

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。

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        int index = Arrays.binarySearch(nums,target);
        if(index < 0) return new int[]{-1,-1};
        int s = index - 1,e = index + 1;
        while(s >= 0 && nums[s] == nums[index]) s--;
        while(e < nums.length && nums[e] == nums[index]) e++;
        return new int[]{++s,--e};
    }
}