【leetcode】423. Reconstruct Original Digits from English【M】【95】

Given a non-empty string containing an out-of-order English representation of digits 

0-9

,
output the digits in ascending order.

Note:

Input contains only lowercase English letters.
Input is guaranteed to be valid and can be transformed to its original digits. That means invalid inputs such as "abc" or "zerone" are not permitted.
Input length is less than 50,000.

Example 1:

Input: "owoztneoer"

Output: "012"

Example 2:

Input: "fviefuro"

Output: "45"

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最开始是用暴力算法，挨个数字判断，是不是在里面，结果超时了

那就只能采用奸诈的方法，观察数字的特殊性，比如有z的一定是zero，有x的一定是six，然后挨个排除，发现根据十个字母就都能计算出来了

计算的过程如下所示

class Solution(object):

def originalDigits(self, s):
res = ''

z = s.count('z')
x = s.count('x')
g = s.count('g')
h = s.count('h')
t = s.count('t')
ss = s.count('s')
v = s.count('v')
f = s.count('f')
i = s.count('i')
o = s.count('o')
print z,x,g,h,t,ss,v,f,i,o

res += '0' * z
res += '1' *(o-(f-(v-ss+x))-(t-h)-z)
res += '2' * (t-h)
res += '3' * (h-g)
res += '4' * (f-(v-ss+x))
res += '5' * (v-ss+x)
res += '6' * x
res += '7' * (ss-x)
res += '8' * g
res += '9' * (i-(v-ss+x)-g-x)

#res.sort()

return res

'''
def check(self,s,num):

for i in num:
if i not in s:
return False
return True
tnums = ['zero','one','two','three','four','five','six','seven','eight','nine']
nums = []
for i in tnums:
i = i.decode()
i = list(i)
#print i
nums += i,
#print nums

s = list(s)

#s.remove('o')
#print s
res = ''
while s:
#print s
for i in nums:
if self.check(s,i):
res += str(nums.index(i))
for ii in i:
s.remove(ii)
return res
'''
"""
:type s: str
:rtype: str
"""