【LeetCode】704. Binary Search 解题报告（Python）

原创

负雪明烛 2022-02-11 17:14:02 博主文章分类：LeetCode ©著作权

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作者：负雪明烛

id： fuxuemingzhu

题目描述

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].

题目大意

二分查找某个元素出现的位置。

解题方法

线性查找

这个题目名字叫做二分查找，但是给的测试用例使用10000个，那么完全可以线性查找，代码如下。

class Solution(object):
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        try:
            return nums.index(target)
        except:
            return -1

二分查找

不懂为啥这么简单的题，没人做？

二分查找真是最基本的题目了吧，应该保证一遍就过的。就不多说了。

下面的做法是查找[left, right]闭区间。代码如下：

class Solution:
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        left, right = 0, len(nums) - 1
        while left <= right:
            mid = (left + right) // 2
            if nums[mid] == target:
                return mid
            elif nums[mid] < target:
                left = mid + 1
            else:
                right = mid - 1
        return -1

如果是查找[left, right)左闭右开区间的话，代码如下：

class Solution(object):
    def search(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        N = len(nums)
        left, right = 0, N
        # [0, N)
        while left < right:
            mid = left + (right - left) / 2
            if nums[mid] == target:
                return mid
            elif nums[mid] > target:
                right = mid
            else:
                left = mid + 1
        return -1