6、Search in Rotated Sorted Array

Problem Description

Suppose a sorted array 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).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Analyze

二分查找法：
如果中间元素正好是要查找的元素，则搜素过程结束；
如果某一特定元素大于或者小于中间元素，则在数组大于或小于中间元素的那一半中查找，而且跟开始一样从中间元素开始比较。

在A[mid] != target情况下：
若A[mid] < A[end]，mid右边升序
若A[mid] > A[end]，mid左边升序
时间复杂度: O(logn)

Code

class Solution {
    func search(nums: [Int], _ target: Int) -> Int {
       var start = 0, end = nums.count - 1
        
        while (start <= end) {
            let mid = start + (end - start) / 2
            
            if nums[mid] == target { return mid }
            
            if nums[mid] < nums[end] { // right half sorted
                if target  > nums[mid] && target <= nums[end] {
                    start = mid + 1
                }
                else { end = mid - 1}
            }
            else { // left half sorted
                if target >= nums[start] && target < nums[mid] {
                    end = mid - 1
                }
                else { start = mid + 1 }
            }
        }
        
        return -1
    }
}

Search in Rotated Sorted Array II

Problem Description

Follow up for "Search in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

Analyze

若A[mid] == A[end] （!= target），无法判定右边搜寻区间为[start , end-1]

Code

class Solution {
    func search(nums: [Int], _ target: Int) -> Bool {
        var start = 0, end = nums.count - 1
        
        while (start <= end) {
            let mid = start + (end - start) / 2
            
            if nums[mid] == target { return true }
            
            if nums[mid] < nums[end] {
                if target  > nums[mid] && target <= nums[end] {
                    start = mid + 1
                }
                else { end = mid - 1}
            }
            else if nums[mid] > nums[end] {
                if target >= nums[start] && target < nums[mid] {
                    end = mid - 1
                }
                else { start = mid + 1 }
            }
            else { end -= 1 }
        }
        
        return false
    }
}