81 Search in Rotated Sorted Array II 搜索旋转排序数组 II
Description:
Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).
You are given a target value to search. If found in the array return true, otherwise return false.
Example:
Example 1:
Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true
Example 2:
Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false
Follow up:
This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
Would this affect the run-time complexity? How and why?
题目描述:
假设按照升序排序的数组在预先未知的某个点上进行了旋转。
( 例如，数组 [0,0,1,2,2,5,6] 可能变为 [2,5,6,0,0,1,2] )。
编写一个函数来判断给定的目标值是否存在于数组中。若存在返回 true，否则返回 false。
示例 :
示例 1:
输入: nums = [2,5,6,0,0,1,2], target = 0
输出: true
示例 2:
输入: nums = [2,5,6,0,0,1,2], target = 3
输出: false
进阶:
这是 搜索旋转排序数组 的延伸题目，本题中的 nums 可能包含重复元素。
这会影响到程序的时间复杂度吗？会有怎样的影响，为什么？
思路:
参考LeetCode #33 Search in Rotated Sorted Array 搜索旋转排序数组
增加两行去重
平均时间复杂度 O(lgn), 最差时间复杂度 O(n), 当数组全是重复元素时需要逐个检查
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
bool search(vector& nums, int target)
{
int low = 0, high = nums.size() - 1;
while (low <= high)
{
while (low < high and nums[high - 1] == nums[high]) --high;
while (low < high and nums[low + 1] == nums[low]) ++low;
int mid = ((high - low) >> 1) + low;
if (nums[mid] == target) return true;
else if (nums[mid] < nums[high])
{
if (nums[mid] < target and target <= nums[high]) low = mid + 1;
else high = mid - 1;
}
else
{
if (nums[mid] > target and target >= nums[low]) high = mid - 1;
else low = mid + 1;
}
}
return false;
}
};
Java:
class Solution {
public boolean search(int[] nums, int target) {
int low = 0, high = nums.length - 1;
while (low <= high) {
while (low < high && nums[high - 1] == nums[high]) --high;
while (low < high && nums[low + 1] == nums[low]) ++low;
int mid = ((high - low) >>> 1) + low;
if (nums[mid] == target) return true;
else if (nums[mid] < nums[high]) {
if (nums[mid] < target && target <= nums[high]) low = mid + 1;
else high = mid - 1;
}
else {
if (nums[mid] > target && target >= nums[low]) high = mid - 1;
else low = mid + 1;
}
}
return false;
}
}
Python:
class Solution:
def search(self, nums: List[int], target: int) -> bool:
return target in nums