[LeetCode] 33. Search in Rotated Sorted Array（搜索旋转有序数组）

• Difficulty: Medium

• Related Topics: Array, Binary Search

• Link: https://leetcode.com/problems/search-in-rotated-sorted-array/

Description

You are given an integer array nums sorted in ascending order, and an integer target.
给定一个升序排序的整数数组 nums，和另一个整数 target。

Suppose that nums is rotated at some pivot unknown to you beforehand (i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).
假设之前 nums 在某处你不知道的地方发生了旋转（也就是说，[0, 1, 2, 4, 5, 6, 7] 可能会变成 [4, 5, 6, 7, 0, 1, 2]）。

If target is found in the array return its index, otherwise, return -1.
如果 target 存在于 nums 中，返回其下标，否则返回 -1。

Examples

Example 1

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Example 3

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

Constraints

• 1 <= nums.length <= 5000
• -10^4 <= nums[i] <= 10^4
• All values of nums are unique.
• nums is guranteed to be rotated at some pivot.
• -10^4 <= target <= 10^4

Solution

这题的关键，在于摆脱这个旋转数组的限制使用二分搜索。以下解法来自 discussion，算是个人认为比较好理解的一种方法了。该方法的核心在于两次二分查找，第一次找到数组发生旋转的那个点，第二次执行常规的二分搜索，具体解法见下：

class Solution {
    fun search(nums: IntArray, target: Int): Int {
        var left = 0
        var right = nums.lastIndex

        // 第一步：找到旋转的点
        while (left < right) {
            val mid = left + (right - left) / 2
            if (nums[mid] > nums[right]) {
                left = mid + 1
            } else {
                right = mid
            }
        }

        val pivot = left
        left = 0
        right = nums.lastIndex
        // 找到 pivot 后，剩下的为常规的二分搜索
        while (left <= right) {
            val mid = left + (right - left) / 2
            val realMid = (mid + pivot) % nums.size
            when {
                nums[realMid] == target -> return realMid
                nums[realMid] < target -> left = mid + 1
                else -> right = mid - 1
            }
        }
        return -1
    }
}