leetcode 34.在排序数组中查找元素的第一个和最后一个位置 Java

题目链接

https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/

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。

如果数组中不存在目标值 target，返回 [-1, -1]。

进阶：

你可以设计并实现时间复杂度为 O(log n) 的算法解决此问题吗？
 
提示：

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

示例

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]

初始代码模板

class Solution {
    public int[] searchRange(int[] nums, int target) {
       
    }
}

代码

去看liweiwei的题解吧，二分如何减少搜索区间：
https://leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array/solution/si-lu-hen-jian-dan-xi-jie-fei-mo-gui-de-er-fen-cha/

还有二分的总结：
https://leetcode-cn.com/problems/search-insert-position/solution/te-bie-hao-yong-de-er-fen-cha-fa-fa-mo-ban-python-/

class Solution {
    public int[] searchRange(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return new int[]{-1, -1};
        }

        int left = findLeftPosition(nums, target);
        if (left == -1) {
            return new int[]{-1, -1};
        }

        int right = findRightPosition(nums, target);

        return new int[]{left, right};
    }

    private int findLeftPosition(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;

        while (left < right) {
            int mid = left + (right - left) / 2;

            if (nums[mid] < target) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }

        if (nums[left] == target) {
            return left;
        }
        return -1;
    }

    private int findRightPosition(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;

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

        return left;
    }
}