704. Binary Search 二分查找
2022年3月7日
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/
@TOC
题目地址:https://leetcode.com/problems/binary-search/description/
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
日期
2018 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