找到的最小数目大于一个给定数量中排序列表 [英] Find the smallest number that is greater than a given number in a sorted list

问题描述

由于数字的排序列表，我需要找到的最小数目大于给定数。考虑一下这个列表：

Given a sorted list of numbers, I need to find the smallest number that is greater than a given number. Consider this list:

arr=[1,2,3,5,7,11,101,131,151,181,191,313,353,373,383]

说出指定的号码是320。然后，我的方法应该返回353为353比320的最小数量越多。

Say the specified number is 320. Then, my method should return 353 as 353 is the smallest number greater than 320.

我想使用二进制搜索稍加修改的形式;然而在执行程序进入无限循环。

I am trying to use a slightly modified form of binary search; however on execution the program goes into infinite loop.

def modBinarySearch(arr,x):
    l=len(arr)
    mid=l/2
    if arr[mid]>=x and arr[mid-1]<x:
        return arr[mid]
    elif arr[mid]>x and arr[mid-1]>x:
        modBinarySearch(arr[mid:l],x)
    else: 
        modBinarySearch(arr[0:mid],x)

N=int(raw_input())
arr=[1,2,3,5,7,11,101,131,151,181,191,313,353,373,383]
print modBinarySearch(arr,N)

有人能指出我在做什么错了？

Can someone point out what I am doing wrong ?

推荐答案

如果ARR [MID]和ARR [中期1]，两者都大于你的号码，你应该在ARR搜索[0：MID]，不要你不觉得？

If arr[mid] and arr[mid-1], both are greater than your number, you should search in arr[0:mid], don't you think?

 elif arr[mid]>x and arr[mid-1]>x:
    modBinarySearch(arr[0:mid],x)
else: 
    modBinarySearch(arr[mid:1],x)

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