[LeetCode] 34. Find First and Last Position of Elements in Sorted Array（在有序数组中寻找某个元素第一次和最后一次出现的位置）

• Difficulty: Medium

• Related Topics: Array, Binary Search

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

Description

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.
给定一个升序排序整数数组 nums，找给定值 target 的第一次和最后一次出现的位置。

If target is not found in the array, return [-1, -1].
如果 target 未找到，返回 [-1, -1]。

Follow up

Could you write an algorithm with O(log n) runtime complexity?
你能在 O(log n) 时间复杂度下完成吗？

Examples

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]

Example 3

Input: nums = [], target = 0
Output: [-1,-1]

Constraints

• 0 <= nums.length <= 1e5
• -1e9 <= nums[i] <= 1e9
• nums is a non-decreasing array.
• -1e9 <= target <= 1e9

Solution

这题实际上是两个子问题的缝合：寻找一个数在数组里的左边界和右边界。这是二分查找的变式，具体解法可以参考这篇博客：详解二分查找算法 - murphy_gb - 博客园 (cnblogs.com)，代码如下：

class Solution {
    fun searchRange(nums: IntArray, target: Int): IntArray {
        return intArrayOf(leftBound(nums, target), rightBound(nums, target))
    }

    private fun leftBound(nums: IntArray, target: Int): Int {
        if (nums.isEmpty()) {
            return -1
        }
        var left = 0
        var right = nums.size
        while (left < right) {
            val mid = left + (right - left) / 2
            if (nums[mid] < target) {
                left = mid + 1
            } else {
                right = mid
            }
        }
        if (left == nums.size) {
            return -1
        }
        return if (nums[left] == target) {
            left
        } else {
            -1
        }
    }

    private fun rightBound(nums: IntArray, target: Int): Int {
        if (nums.isEmpty()) {
            return -1
        }
        var left = 0
        var right = nums.size
        while (left < right) {
            val mid = left + (right - left) / 2
            if (nums[mid] <= target) {
                left = mid + 1
            } else {
                right = mid
            }
        }
        if (left == 0) {
            return -1
        }
        return if (nums[left - 1] == target) {
            left - 1
        } else {
            -1
        }
    }
}