Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.
You must write an algorithm with O(log n) runtime complexity.
規則
給予一個數字 target 和一個數組 nums，如果 nums 中存在 target，則返回 target 的索引，反之返回 -1。
測試案例
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
思路
- 取出中間值
- 如果 low > high ,則代表沒找到，回傳 -1
- 如果中間值等於 target，命中則回傳
- 如果中間值大於 target，則從中間值開始往左找
- 如果中間值小於 target，則從中間值開始往右找
實作
def search(self, nums, target) -> int:
return self.helper(nums, 0, len(nums) - 1, target)
def helper(self, nums, low, high, target):
mid = (low + high) // 2 # 取整數
if low > high: # 沒找到
return -1
if nums[mid] == target: # 命中
return mid
if nums[mid] > target: # 從左邊找
high = mid - 1 # 從中間值開始往左找
if nums[mid] < target: # 從右邊找
low = mid + 1 # 從中間值開始往右找
return self.helper(nums, low, high, target)