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:

  1. You may assume that all elements in nums are unique.
  2. n will be in the range [1, 10000].
  3. 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.

References

  1. 704. Binary Search