[LeetCode] 33. 搜索旋转排序数组
题目链接: https://leetcode-cn.com/problems/search-in-rotated-sorted-array/
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 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。
搜索一个给定的目标值，如果数组中存在这个目标值，则返回它的索引，否则返回 -1 。
你可以假设数组中不存在重复的元素。
你的算法时间复杂度必须是 O(log n) 级别。
示例:
示例 1:
输入: nums = [4,5,6,7,0,1,2], target = 0
输出: 4
示例 2:
输入: nums = [4,5,6,7,0,1,2], target = 3
输出: -1
思路:
题目中说明要用O(log n)能想到就是二分法!
思路一:
整体思路:先用二分法找出最小值,也是那个分割点,例如[4,5,6,7,0,1,2],我们找出数字0;
接下来判断target是在分割点的左边还是右边;
最后再使用一次二分法找出target的位置. 所以时间复杂度为:\(O(logn)\)
只有一个难点,那就是如何通过二分法找出那个分割点呢?
就是和它的右端点比较判断,直接看代码吧!
思路二:
直接使用二分法,
判断那个二分点,有几种可能性
-
直接等于
target -
在左半边的递增区域
a.
target在left和mid之间b.不在之间
-
在右半边的递增区域
a.
target在mid和right之间b. 不在之间
思路三:
还是很容易理解的!
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代码:
方法一:
python
class Solution:
def search(self, nums: List[int], target: int) -> int:
if not nums:return -1
n = len(nums)
left = 0
right = len(nums) - 1
while left < right:
mid = left + (right - left) //2
if nums[mid] > nums[right]:
left = mid + 1
else:
right = mid
t = left
left = 0
right = len(nums) - 1
while left <= right:
mid = (left + right) //2
realmid = (mid + t) % n
if nums[realmid] == target:
return realmid
elif nums[realmid] > target:
right = mid - 1
else:
left = mid + 1
return -1
java
class Solution {
public int search(int[] nums, int target) {
if (nums == null || nums.length == 0) return -1;
int left = 0;
int right = nums.length - 1;
while (left < right) {
int mid = left + (right - left) / 2;
if (nums[mid] > nums[right]) left = mid + 1;
else right = mid;
}
//System.out.println(left);
int split_t = left;
left = 0;
right = nums.length - 1;
if (nums[split_t] <= target && target <= nums[right]) left = split_t;
else right = split_t;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) return mid;
else if (nums[mid] > target) right = mid - 1;
else left = mid + 1;
}
return -1;
}
}
方法二
python
class Solution:
def search(self, nums: List[int], target: int) -> int:
n = len(nums)
if n == 0:
return -1
left = 0
right = n - 1
while left < right:
mid = left + (right - left) // 2
if nums[mid] == target:
return mid
elif nums[left] <= nums[mid]:
if nums[left] <= target < nums[mid]:
right = mid - 1
else:
left = mid + 1
else:
if nums[mid] < target <= nums[right]:
left = mid + 1
else:
right = mid - 1
#print(left,right)
return left if nums[left] == target else -1
java
class Solution {
public int search(int[] nums, int target) {
if (nums == null || nums.length == 0) return -1;
int n = nums.length;
int left = 0;
int right = n - 1;
while (left < right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) return mid;
else if (nums[left] <= nums[mid]) {
if (nums[left] <= target && target < nums[mid]) right = mid - 1;
else left = mid + 1;
} else if (nums[mid] < nums[right]) {
if (nums[mid] < target && target <= nums[right]) left = mid + 1;
else right = mid - 1;
}
}
return nums[left] == target ? left : -1;
}
}
思路三:
python
class Solution:
def search(self, nums: List[int], target: int) -> int:
n = len(nums)
if n == 0:
return -1
left = 0
right = n
while left < right:
mid = left + (right - left) // 2
# if (target < nums[0] and nums[mid] < nums[0]) or \
# (target > nums[0] and nums[mid] > nums[0])
if (nums[mid] < nums[0]) == (target < nums[0]):
num = nums[mid]
else:
if target < nums[0]:
num = float("-inf")
else:
num = float("inf")
if num < target:
left = mid + 1
elif num > target:
right = mid
else:
return mid
return -1
java
class Solution {
public int search(int[] nums, int target) {
if (nums == null || nums.length == 0) return -1;
int left = 0;
int right = nums.length;
while (left < right) {
int mid = left + (right - left) / 2;
long num = (nums[mid] < nums[0]) == (target < nums[0])
? nums[mid]
: target < nums[0] ? Long.MIN_VALUE: Long.MAX_VALUE;
if (num < target) left = mid + 1;
else if (num > target) right = mid;
else return mid;
}
return -1;
}
}
