2019/8/27 滴滴算法笔试题1

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最新推荐文章于 2025-03-09 18:53:29 发布

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分类专栏： python 笔试

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本文介绍了一种针对包含加、减、乘、除运算的表达式进行顺序优化的算法，该算法通过重新排列相邻数字以达到不改变运算结果的同时，获得更优的运算顺序组合。示例中，输入表达式为'3+2+1*6*-2/5'，输出优化后的顺序为'2+3+-2*1*6/5'。

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# -*- coding: utf-8 -*-
"""
Created on Wed Aug 28 09:51:02 2019

@author: QinLong
给定一组四则运算（+ - * /）,只能相邻数字相交换，求不改变运算结果的最小顺序组合。
eg:
input: 3 + 2 + 1 * 6 * -2 / 5
out : 2 + 3 + -2 * 1 * 6 / 5
"""

  
        
number = [3,2,1,6,-2,5]
n = len(number)
fuhao = ['+','+','*','*','/']
       
res = []
count = 0       
for j in range(len(fuhao)):
    if fuhao[j] == '*':
        
        count = j
        count_2 = 1
        while count < n-2 and fuhao[count + 1] == '*':
            count += 1
            count_2 += 1
        if j > 0:     
            if fuhao[j-1] == '/':
                j +=1
            
        tem = number[j:count+2]
        tem.sort()
        for k in range(len(tem)):
            number[j+k] = tem[k]
        
        print(number)
    elif fuhao[j] == '/' :
        count = j
        count_2 = 1
        while count < n-2 and fuhao[count + 1] == '/':
            count += 1
            count_2 += 1
        if count_2 > 1:

            tem = number[j+1:count+2]
            tem.sort()
            for k in range(len(tem)):
                number[j+k+1] = tem[k]
                
        print(number)
    elif fuhao[j] == '+':
        count = j
        count_2 = 1
        while count < n-2 and fuhao[count + 1] == '+':
            count += 1
            count_2 += 1
        if j > 0:  
            if fuhao[j-1] == '*' or fuhao[j-1] == '/':
                j +=1  
        if count <n-2: 
            if fuhao[count+1] == '*' or fuhao[count+1] == '/':
                count -= 1

        tem = number[j:count+2]
        tem.sort()
        for k in range(len(tem)):
            number[j+k] = tem[k]  
        print(number)
    else:
        count = j
        count_2 = 1
        while count < n-2 and fuhao[count + 1] == '-':
            count += 1
            count_2 += 1
        if j > 0:    
            if fuhao[j-1] == '*' or fuhao[j-1] == '/':
                j +=1 
        if count <n-2:     
            if fuhao[count + 1] == '*' or fuhao[count + 1] == '/':
                count -= 1
        
        if count_2 > 1:

            tem = number[j+1:count+2]
            tem.sort()
            for k in range(len(tem)):
                number[j+k+1] = tem[k]
        print(number)    
for f in range(n-1):
    res.append(number[f])
    res.append(fuhao[f])           
res.append(number[-1])

for r in res:
    print(r, end=' ')