博客链接: http://codeshold.com/2017/03/alialgorithm.html
题目
2017年3月阿里在线编程题(实习内推)
给定一串数字
判断是否存在这三个元素，它们将数字串分为四个子串，其中每个子串的数字之和均相同(该3个元素不纳入计算)
要求时间复杂度和空间复杂度均不能超过O(n)
实现
python 写的
代码不够优美，欢迎留言指正！
注：添加了测试代码，所以比较冗长!
# -*- encoding:utf-8 -*-
class StrSplit(object):
def __init__(self, srclist):
self.srclist = srclist
self.sumA = [0] * len(self.srclist) # 保存从左到右依次累积统计的sum
self.sumB = [0] * len(self.srclist) # 保存从右到左依次累积统计的sum
self.dictA = {} # 保存sumA中具有某个值的索引号
# 如dictA[100]为[1,12],即表示sumA[1]和sum[12]都是100
self.dictB = {} # 保存sumB中具有某个相同索引号
self.result = [0, 0, 0] # 保存统计结果，即需要删除的三个元素的索引
self.initstrsplit()
def initstrsplit(self):
sum = 0
for i in range(0, len(self.srclist)):
sum += self.srclist[i]
self.sumA[i] = sum
if self.sumA[i] in self.dictA:
self.dictA[self.sumA[i]].append(i)
else:
self.dictA[self.sumA[i]] = [i]
sum = 0
for i in range(len(self.srclist)-1, -1, -1):
sum += self.srclist[i]
self.sumB[i] = sum
if self.sumB[i] in self.dictB:
self.dictB[self.sumB[i]].append(i)
else:
self.dictB[self.sumB[i]] = [i]
def getmidpivot(self, left, right, partsum):
if not (1 < left < right < len(self.srclist)-2 and left < right - 1):
return False
#print "left:%d, right:%d, partsum:%d" %(left, right, partsum)
mid = self.sumA[left-1] + partsum # left-1为要删除的第一个元素
if mid in self.dictA:
for i in self.dictA[mid]: # i+1 为要删除的中间元素
if left-1 < i < right-1:
#print self.sumA[right], self.sumA[i]
if self.sumA[right] - self.sumA[i+1] == partsum:
return i+1
else:
continue
return False
def getresult(self):
if len(self.srclist) < 7:
return False
for i in range(0, len(self.sumA)-4):
if self.sumA[i] in self.dictB:
index = self.dictB[self.sumA[i]] # index is list type
for j in index:
if j > i+3:
# 删除元素i+1 和 j-1后, 判断是否存在中间元素
m = self.getmidpivot(i+2, j-2, self.sumA[i])
if m:
self.result = [i+1, m, j-1]
return True
return False
def checkresult(self):
sum1, sum2, sum3, sum4 = 0, 0, 0, 0
left, mid, right = tuple(self.result)
for i in range(0, left):
sum1 += self.srclist[i];
for i in range(left+1, mid):
sum2 += self.srclist[i];
if sum2 != sum1: return False
for i in range(mid+1, right):
sum3 += self.srclist[i];
if sum3 != sum1: return False
for i in range(right+1, len(self.srclist)):
sum4 += self.srclist[i];
if sum4 != sum1: return False
return True
if __name__ == "__main__":
import random
a = [1,2,3,4,5,6,7,8,9,10,1,2,3,24,3,4,1,3,4,5,3,2,1,2,3,4,5,6,7]
b = [1,1,1,9,1,2,3,-3,4,1,2,5,1,2,-100,100,2,3,-5]
c = [1,1,1,1,1,1,1]
d = []
e = [0,0,0,0,0,0,0,0,0,0]
all = [a, b, c, d, e]
for i in range(0, 100): # 列表数
tmp = []
for j in range(0, random.randint(1, 1000)): # 元素的个数
tmp.append(random.randint(-10, 10)) # 元素的范围
all.append(tmp)
okcount = 0
failcount = 0
errorcount = 0
for i in all:
swf = StrSplit(i)
# print swf.srclist
ret = swf.getresult()
if ret:
okcount += 1
if not swf.checkresult(): errorcount += 1
print " StrSplit: True "
print " Result: ", swf.result
print " Check result: ", swf.checkresult()
else:
failcount += 1
print " SUCCESS COUNT: ", okcount
print " FAILED COUNT: ", failcount
print " ERROR COUNT: ", errorcount
