Find First and Last Position of Element in Sorted Array

找有重复数字的左边界：

1. mid >= target end= mid ,不断缩小右窗口，

2. 最后check start和end的时候，先check start，因为要取左窗口

找有重复数字的右边界：

1. mid <= target start = mid 不断缩小左窗口

2. 最后check start和end的时候，先check end，因为要取右窗口

Given an array of integers nums 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].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

class Solution {
    public int[] searchRange(int[] nums, int target) {
        
        int[] res = {-1,-1};
        
        if(nums == null || nums.length == 0)
            return res;
        
        
        res[0] = findLeft(nums,target);
        res[1] = findRight(nums,target);
     
        return res;
        
    }
    
    public int findLeft(int[] nums, int target){
        int start = 0, end = nums.length-1;
        
        while(start + 1 < end){
            int mid = start + (end - start)/2;
            if(nums[mid] >= target){
                end = mid;
            }else{
                start = mid;
            }
        }
        
        if(nums[start] == target){
            return start;
        }
        else if(nums[end] == target){
            return end;
        }else{
            return -1;
        }
    }
    
    public int findRight(int[] nums,int target){
         int start = 0, end = nums.length-1;
        
        while(start + 1 < end){
            int mid = start + (end - start)/2;
            if(nums[mid] <= target){
                start = mid;
            }else{
                end = mid;
            }
        }
        
        if(nums[end] == target){
            return end;
        }
        else if(nums[start] == target){
            return start;
        }else{
            return -1;
        }
    }
}