在排序数组中查找元素的第一个和最后一个位置

题目描述：
给定一个按照升序排列的整数数组 nums，和一个目标值 target。找出给定目标值在数组中的开始位置和结束位置。

你的算法时间复杂度必须是 O(log n) 级别。

如果数组中不存在目标值，返回 [-1, -1]。

示例：
输入: nums = [5,7,7,8,8,10], target = 8
输出: [3,4]

Java代码：
大神链接[https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/solution/er-fen-cha-zhao-suan-fa-xi-jie-xiang-jie-by-labula/]

class Solution {
    public int binary_search(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        int mid;

        while(left <= right) {
            mid = left + (right - left) / 2;
            if(target == nums[mid]) return mid;
            else if(target < nums[mid]) right = mid - 1;
            else if(target > nums[mid]) left = mid + 1;
        }

        return -1;
    }

    public int left_bound(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        int mid;

        while(left <= right) {
            mid = left + (right - left) / 2;
            if(target == nums[mid]) right = mid - 1;
            else if(target < nums[mid]) right = mid - 1;
            else if(target > nums[mid]) left = mid + 1;
        } 

        if(left >= nums.length || nums[left] != target) return -1;
        return left;
    }

    public int right_bound(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        int mid;

        while(left <= right) {
            mid = left + (right - left) / 2;
            if(target == nums[mid]) left = mid + 1;
            else if(target < nums[mid]) right = mid - 1;
            else if(target > nums[mid]) left = mid + 1;
        }

        if(right < 0 || nums[right] != target) return -1;
        return right;
    }

    public int[] searchRange(int[] nums, int target) {
        if(nums.length == 0) return new int[]{-1,-1};
        int left = left_bound(nums, target);
        int right = right_bound(nums, target);
        return new int[]{left, right};
    }
}