传闻中的华为Python面试题（原创）

原题:
有两个序列a,b，大小都为n,序列元素的值任意整形数，无序；
要求：通过交换a,b中的元素，使[序列a元素的和]与[序列b元素的和]之间的差最小

限时十分钟以内

 

尝试解答，coding和debug共用了约1个小时，呵呵，去不了华为了。
获取最小差的函数，还可以优化，现在的时间复杂度是n×n，估计可以优化到n×lg(n)

 

网上有一种解答方法，认为最小值是负数，排序之后，截成两段，小数序列-大数序列即可。难道，难道，这才是本题十分钟解出的奥义？！

我的代码如下，算法很简单（我的算法不对，请列位看官参考评论中张三的）：

 

 

'''
Created on 2010-6-21

@author: maodouzi
'''
import random

def getMinSubKeys(leftArray, rightArray):
    if (sum(leftArray) > sum(rightArray)):
        subArray = leftArray
        minuendArray = rightArray
        subFlag = True
    elif (sum(leftArray) < sum(rightArray)):
        subArray = rightArray
        minuendArray = leftArray
        subFlag = False
    else:
        return None
    
    minSubNum = (sum(subArray) - sum(minuendArray)) / 2.0
    
    tmpMinSubResult = sum(subArray)
    subKey = 0
    minuendKey = 0
    for i_key, i in enumerate(subArray):
        for j_key, j in enumerate(minuendArray):
            if (abs(i - j - minSubNum) < tmpMinSubResult):
                tmpMinSubResult = abs(i - j - minSubNum)
                subKey = i_key
                minuendKey = j_key
                
    if (subFlag == True):
        return (subKey, minuendKey)
    else:
        return (minuendKey, subKey)

if __name__ == '__main__':
    aa = range(20);
    bb = [random.randint(0, 100) for i in aa]
    bb.sort(reverse=True)
    print aa
    print bb
    
    leftArray = []
    rightArray = []
    while (len(bb)):
        tmpNumArray = bb[0:2]
        bb = bb[2:]
        
        if (sum(leftArray) <= sum(rightArray)):
            leftArray.append(tmpNumArray[0])
            rightArray.append(tmpNumArray[1])
        else :
            leftArray.append(tmpNumArray[1])
            rightArray.append(tmpNumArray[0])
            
        leftArray.sort()
        rightArray.sort()
        tmpKeyArray = getMinSubKeys(leftArray, rightArray)
        if (tmpKeyArray == None):
            continue
        else:
            orgSubNum = abs(sum(leftArray) - sum(rightArray));
            newSubNum = abs(orgSubNum + (rightArray[tmpKeyArray[1]] - leftArray[tmpKeyArray[0]]) * 2)
            if (newSubNum < orgSubNum):
                tmpNum = leftArray[tmpKeyArray[0]]
                leftArray[tmpKeyArray[0]] = rightArray[tmpKeyArray[1]]
                rightArray[tmpKeyArray[1]] = tmpNum
        
        leftArray.sort()
        rightArray.sort()
    
    print leftArray, sum(leftArray)
    print rightArray, sum(rightArray)
    
    print "The subNum is: %d" % abs(sum(leftArray) - sum(rightArray))

 

 

输出结果如下：

 

 

[93, 91, 90, 82, 81, 74, 74, 74, 74, 68, 60, 57, 49, 48, 48, 45, 36, 35, 29, 22]
[22, 36, 48, 48, 60, 68, 74, 81, 82, 93] 612
[29, 35, 45, 49, 57, 74, 74, 74, 90, 91] 618
The subNum is: 6