把一个list分成二个list,使其差最小

[liu510817387]

这是在checkio.com上的一道题，初学python,就采用最笨的方法
list拆分成二个差最小的list
例如：[5,10,5,10,1] => sum([5,10,1]) - sum([5,10]) = 1

def checkio(data):
    dataLen = len(data)
    '''data.sort()'''
    if dataLen==1:
        return data[0]
    listDiff = []
    for i in range(dataLen-1,-1,-1):
        leftList = [data[i]]
        rightList = data[:]
        del rightList[i]
        for y in range(dataLen-2,-1,-1):
            if rightList[y]:
                addItem = rightList[y]
                leftSum = sum(leftList)
                rightSum = sum(rightList)
                if leftSum+2*addItem<=rightSum:
                    leftList.append(addItem)
                    del rightList[y]
        leftSum = sum(leftList)
        rightSum = sum(rightList)
        listDiff.append(abs(leftSum-rightSum))
    listDiff.sort()
    result = listDiff[0]
    #replace this for solution
    return result

不知道这样写为什么错了？
[5,10,5,10,1] 测试这个时候出错了
（PS:欢迎提供动态规划版的或贪心算法版的）

[liu510817387]

是因为我没有表达清楚吗？

[梅小西Echo]

a=[3,56,78,1,5,9,90,39,20,73,91,30,26]


def split(alist):
    alist.sort()
    alist.reverse()
    resultA=[]
    resultB=[]
    for i in alist:
        if sum(resultA)>sum(resultB):
            resultB.append(i)
        else:
            resultA.append(i)
    return resultA,resultB,sum(resultA)-sum(resultB)
print(split(a))

[libralibra]

2楼的算法在a=[12, 30, 30, 32, 42, 49]时的结果(13)有误,应该是9 下面是我用01背包解决的代码,思路是: 总数据A求和S后,问题转化为:用A中的数据填充容量为S/2的背包,得到最大价值V.那么原问题2部分元素的和的差就是: |(S-V)-V|,由于0-1背包问题时背包容量是S/2,所以可知S-V肯定不小于V,因此最后的差值是:S-2V.

def checkio(data):
    sd = sum(data)      # all sum
    std = sorted(data)  # sort items
    sm = chk(sd/2,std)  # find the biggest value
    return sd-sm-sm     # find the max_half-min_half

# 0-1 knapsack solution
def chk(v,s):           # value, item
    if len(s)==0 or v<s[0]: # no item or the smallest cannot be put into bag
        return 0
    else:               # some can be put into bag
        pv = []
        for i in range(len(s)):
            if s[i]<=v:
                cv = max(s[i]+chk(v-s[i],s[:i]+s[i+1:]),chk(v,s[:i]+s[i+1:]))
                pv.append(cv)
        if not pv:
            return 0
        else:
            return max(pv)

[liu510817387]

#3 楼的是正确的 thx