Leetcode35 Search Insert Position 解题思路(python)

本人编程小白,如果有写的不对、或者能更完善的地方请个位批评指正!

这个是leetcode的第35题,这道题的tag是数组,python里面叫list,需要用到二分搜索法

35. Search Insert Position

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

Example 1:

Input: [1,3,5,6], 5Output: 2

Example 2:

Input: [1,3,5,6], 2Output: 1

Example 3:

Input: [1,3,5,6], 7Output: 4

Example 4:

Input: [1,3,5,6], 0Output: 0

(https://leetcode.com/problems/search-insert-position/)

思路:

方法一:

这道题目最直观的解法肯定是一次循环for循环,因为数组(list)已经是排好序得了,考虑两种情况:

第一种:target不在数组中,那么比数组中最大的数字大的时候,他的返回值是数组的长度+1(即python中的len(array)),比数组中最小的数字小的在第二种情况中考虑,返回值是0

第二种:只要找出来第一个比target大的数字,那么这个数字所对应的位置就是返回值,

时间复杂度:O(n)

Python 代码实现:

class Solution(object):

    def searchInsert(self, nums, target):

        """

        :type nums: List[int]

        :type target: int

        :rtype: int

        """

        if target > nums[len(nums)-1]:

            return len(nums)

        for i in range(len(nums)):

            if nums[i] >= target:

                return i

代码实现起来其实是很容易的,只有五行,但我用的第一种方法做完之后只beat了2.8%的提交者(2018/12/22,因为这个“%”肯定随着时间的推移而改变),说明肯定不是最优解

方法二:

既然是排好序的数组,那么我们知道二分搜索法的时间复杂度是log(n),(如果有人需要复习二分搜索发的话请见参考文献)

类似于方法一的第二种情况,我们是希望找到数组中第一个比target大的数字

这里我考虑的是以下几种情况:

第一种:target小于num[0]或者大于num[len-1]

第二种:target等于数组中的某数,即被二分搜索法找到

第三种,target在num[0]和num[len-1]中间,但不等于数组中的任何一个数

时间复杂度:log(n)

Python 代码实现:

class Solution(object):

    def searchInsert(self, nums, target):

        """

        :type nums: List[int]

        :type target: int

        :rtype: int

        """

        if target <= nums[0]:

            return 0

        elif target > nums[len(nums)-1]:

            return len(nums)

        low,high = 0, len(nums)-1

        while low <= high:

            mid = (low+high) // 2

            if target > nums[mid]:

                low = mid + 1

            elif target < nums[mid]:

                high = mid - 1

            else:

                return mid

        if target < nums[mid]:

            return mid

        elif target > nums[mid]:

            return mid+1

    if __name__ == ‘__main__‘:

        nums = [1,2,3,4,6]

        target = 5

        print(searchInsert(0,nums,target))

我这里用的是while实现的二分搜索法,提交之后发现竟然还只是击败了32%的提交者,但感觉在时间复杂度上已经是最优的解了,

如果还有更好的解法欢迎大家留言或私信,谢谢 : )

参考文献:

1. 复习二分法连接:https://blog.csdn.net/djd1234567/article/details/45676829

原文地址:https://www.cnblogs.com/samstudy/p/10163231.html

时间: 2024-10-12 10:32:23

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