Given a sorted (in ascending order) integer array nums of n elements and a target value, write a function to search target in nums. If target exists, then return its index, otherwise return -1.
Example 1:
Input: nums = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4
Example 2:
Input: nums = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1
Note:
- You may assume that all elements in nums are unique.
- n will be in the range [1, 10000].
- The value of each element in nums will be in the range [-9999, 9999].
Solutions
1. 递归方法
注意判断条件，right 最开始要是 n-1（不然在计算时会出现索引越界），且在判断返回为 -1 时，要选择 left > right（left 和 right 相等是，求得的 mid 还是相同的数，是满足查找条件的）。
# Time Complexity: O(logn)
# Space Complexity: O(1)
class Solution:
def search(self, nums: List[int], target: int) -> int:
if not nums:
return -1
left, right = 0, len(nums)-1
return self.binary_search(nums, left, right, target)
def binary_search(self, nums: List[int], left: int, right: int, target: int) -> int:
if left > right:
return -1
mid = (left + right) >> 1
if nums[mid] == target:
return mid
elif nums[mid] > target:
right = mid - 1
else:
left = mid + 1
return self.binary_search(nums, left, right, target)
# Runtime: 304 ms, faster than 10.61% of Python3 online submissions for Binary Search.
# Memory Usage: 15 MB, less than 6.45% of Python3 online submissions for Binary Search.
2. 迭代方法
迭代相对来说会快一点，没有重复进出调用栈。
# Time Complexity: O(logn)
# Space Complexity: O(1)
class Solution:
def search(self, nums: List[int], target: int) -> int:
if not nums:
return -1
left, right = 0, len(nums)-1
while left <= right:
mid = (left + right) >> 1
if nums[mid] == target:
return mid
elif nums[mid] > target:
right = mid - 1
else:
left = mid + 1
return -1
# Runtime: 292 ms, faster than 41.46% of Python3 online submissions for Binary Search.
# Memory Usage: 15 MB, less than 6.45% of Python3 online submissions for Binary Search.