704. Binary Search 二分查找
作者: 负雪明烛 id: fuxuemingzhu 个人博客: https://fuxuemingzhu.cn/
题目地址:https://leetcode.com/problems/binary-search/description/
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 = 9Output: 4Explanation: 9 exists in nums and its index is 4Example 2:
Input: nums = [-1,0,3,5,9,12], target = 2Output: -1Explanation: 2 does not exist in nums so return -1Note:
- You may assume that all elements in
numsare unique. - n will be in the range
[1, 10000]. - The value of each element in
numswill be in the range[-9999, 9999].
二分查找某个元素出现的位置。
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 -12018 年 7 月 12 日 —— 天阴阴地潮潮,已经连着两天这样了 2018 年 11 月 21 日 —— 又是一个美好的开始

评论与交流