Python - 实现三位数加三位数等于三位数问题（dfs）

算法描述：将1-9放入9个盒子中，使其满足XXX + XXX = XXX，请问有多少种情况（去重）?

算法实现：采用了dfs的核心思想

a = [0] * 9
book = [0] * 9
TOTAL = 0

def dfs(step):
    if (step == len(a)):
        con1 = a[0] * 100 + a[1] * 10 + a[2]
        con2 = a[3] * 100 + a[4] * 10 + a[5]
        con3 = a[6] * 100 + a[7] * 10 + a[8]
        if ((con1 + con2) == con3):
            global TOTAL
            TOTAL += 1
            print('{0}{1}{2} + {3}{4}{5} = {6}{7}{8}'.format(a[0], a[1], a[2], a[3], a[4], a[5], a[6], a[7], a[8]))
        return

    for i in range(1,10):
        if (book[i-1] == 0):  # still in hand
            a[step] = i
            book[i-1] = 1     # not in hand

            dfs(step + 1)     # Recursive to next step
            book[i-1] = 0     # The most important step to take it on hand

if __name__=='__main__':
    dfs(0)
    #print(TOTAL) #一共有336种可能
    print('TOTAL:{0}'.format(TOTAL//2))  #去重过后有168种