34. Search for a Range

Given an array of integers sorted in ascending order, 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].

本题开始确实能够想到二分查找，但是，却并不知道如何取得最左面和最右面值==target的索引，本题的二分查找是查找两次，一次是找最左面的值==target的索引，一次是找最右面的值==target的索引，拿最左面举例，以上题为例，在7，8之间进行划分，也就是说mid>=和mid<分成两种情况，接下来就是二分查找了，代码如下：

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        if(nums==null||nums.length==0) return new int[]{-1,-1};
        int left = 0;
        int right = nums.length-1;
        while(left<right){
            int mid = left +(right-left)/2;
            if(nums[mid]>=target){
                right = mid;
            }else{
                left = mid+1;
            }
        }
        if(nums[left]!=target) return new int[]{-1,-1};
        int left_index = left;
        right = nums.length-1;
        while(left<right){
            int mid = left+(right-left)/2+1;
            if(nums[mid]<=target) left  =mid;
            else{
                right = mid-1;
            }
        }
        int right_index = left;
        return new int[]{left_index,right_index};
    }
}
//the time complexity of this one could be O(logn), the space complexity could be O(1);