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Difficulty Level: Moderate
1. Description
It is known that there exists an array of integers in non-descending order nums, and the values in the array do not have to be different from each other.
Before being passed to the function, a rotation is performed on numssome subscript k( ) unknown in advance , so that the array becomes ( subscript counting from 0). For example, after rotation at the subscript it might become .0 <= k < nums.length[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]][0,1,2,4,4,4,5,6,6,7]5[4,5,6,6,7,0,1,2,4,4]
Given your rotated array numsand an integer target, please write a function to determine whether the given target value exists in the array. Returns if the target value exists numsin , otherwise .targettruefalse
You have to minimize the whole operation steps as much as possible.
2. Examples
Example 1
输入：nums = [2,5,6,0,0,1,2], target = 0
输出：true
Example 2
输入：nums = [2,5,6,0,0,1,2], target = 3
输出：false
Restrictions:
1 <= nums.length <= 5000-10^4 <= nums[i] <= 10^4- The title data is
numsguaranteed to be rotated on a subscript unknown in advance -10^4 <= target <= 10^4
Advanced:
- This is
搜索旋转排序数组an extended topic of , in this topicnumsmay contain repeated elements. - Will this affect the time complexity of the program? What will be the impact and why?
3. Answer
class SearchInRotatedSortedArrayII {
func search(nums: [Int], _ target: Int) -> Bool {
var left = 0
var right = nums.count - 1
var mid = 0
while left <= right {
mid = (right - left) / 2 + left
if nums[mid] == target {
return true
}
if nums[mid] > nums[left] {
if nums[mid] > target && target >= nums[left] {
right = mid - 1
} else {
left = mid + 1
}
} else if nums[mid] < nums[left]{
if nums[mid] < target && target <= nums[right] {
left = mid + 1
} else {
right = mid - 1
}
} else {
left += 1
}
}
return false
}
}
- Main idea: binary search, check if left or right is sorted, then look in the section, jump if duplicate.
- Time complexity: O(logn)
- Space Complexity: O(1)
The warehouse of the algorithm solution: LeetCode-Swift
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