在排序数组中查找元素的第一个和最后一个位置

给你一个按照非递减顺序排列的整数数组 nums，和一个目标值 target。请你找出给定目标值在数组中的开始位置和结束位置。力扣题目

如果数组中不存在目标值 target，返回 [-1, -1]。

你必须设计并实现时间复杂度为 O(log n) 的算法解决此问题。

示例 1：

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

示例 2：

输入： nums = [5,7,7,8,8,10], target = 6
输出： [-1,-1]

示例 3：

输入： nums = [], target = 0
输出： [-1,-1]

提示：

• 0 <= nums.length <= 105
• -109 <= nums[i] <= 109
• nums 是一个非递减数组
• -109 <= target <= 109

解题思路：左边界和右边界分两次求，再合起来返回

public class SearchRange {
    public static void main(String[] args) {
        int nums[] = {5,7,7,8,8,10}; int target = 8;
//        int nums[] = {5,7,7,8,8,10}; int target = 6;
//        int nums[] = {}; int target = 0;
        int[] ints = searchRange(nums, target);
        for (int i = 0; i < ints.length; i++) {
            System.out.print(ints[i]+ "---");
        }
        System.out.println("--------------------------------");
    }

    public static int[] searchRange(int[] nums, int target) {

        int start = searchStart(nums, target);
        int end = searchEnd(nums, target);
        return new int[]{start,end};
    }
    public static int searchStart(int[] nums, int target) {
        int left =0;
        int right =nums.length - 1;
        int start = -1;
        while(left <= right) {
            int middle = left + (right - left) / 2;
            if(nums[middle] == target) {
                start = middle;
                right = middle - 1;
            }else if(nums[middle] <target){
                left = middle + 1;
            }else if(nums[middle] > target) {
                right = middle - 1;
            }
        }
        return start;
    }
    public static int searchEnd(int[] nums, int target) {
        int left =0;
        int right =nums.length - 1;
        int start = -1;
        while(left <= right) {
            int middle = left + (right - left) / 2;
            if(nums[middle] == target) {
                start = middle;
                left = middle + 1;
            }else if(nums[middle] <target){
                left = middle + 1;
            }else if(nums[middle] > target) {
                right = middle - 1;
            }
        }
        return start;
    }
}