全排列序列

合法括号序列

list深度

‘’‘比如这个测试用例：L = [1, 2, 3, [4, [5, 6]],[[[[]]]],[]]应该是5层‘’’cnt=0 rst=0 for i in str(L): if i==&#x27;[&#x27;:         cnt+=1         rst=max(rst,cnt) if i==&#x27;]&#x27;:         cnt-=1 print(rst)

就是比较进出的括号，取最大

<pre><code>def get_res(L):     L=str(L)     op_stack=[]#临时     op_stack_len=[]#记录深度 for i in L: if i==&#x27;[&#x27;:             op_stack.append(i)             op_stack_len.append(len(op_stack)) elif i ==&#x27;]&#x27;:             op_stack.pop()             op_stack_len.append(len(op_stack)) else: pass return max(op_stack_len) print(get_res(L)) </code></pre>

基于栈的思想

[1,[1,2],[2,3],[1,4]],很明显就是错误的

str(L).count('[')

<pre class="prettyprint lang-py linenums">S=str(L)
a=0
List=[]
for Str in S:
 if Str=='[':
 a += 1
 List.append(a)
 elif Str==']':
 a -= 1
 List.append(a)
print(max(List))</pre>

最最简单的方法

<pre class="prettyprint lang-py linenums"># L = [1, 2, 3, [4, [5, 6]]]


# 定义一个新的list，将原list中的子list全部并排放入新list中，循环直到新的list没有子list
def get_sub_list(x):
 n_lst = []
 for y in x:
 if type(y) == list:
 for z in y:
 n_lst.append(z)
 return n_lst


new_lst = L
count = 0
while len(new_lst) &gt; 0:
 count += 1
 new_lst = get_sub_list(new_lst)
print(count)</pre>

不用递归，不用找'['，其实也简单

<pre style="background-color:#2b2b2b;color:#a9b7c6;font-family:'Consolas';font-size:10.5pt;">depth = 1
max = -999999999
floors = ['[']
for i in str(L)[1:]:
 if i=='[':
 floors.append(i)
 depth+=1
 elif i==']':
 depth-=1
 floors.pop()
 if depth &gt; max:
 max = depth

print(max)</pre>

队列

list = [] s = str(L) a = 0 for i in s: &nbsp; &nbsp; if i == &quot;[&quot; : &nbsp; &nbsp; &nbsp; &nbsp; a += 1 &nbsp; &nbsp; &nbsp; &nbsp; list.append(a)s &nbsp; &nbsp; if i == &quot;]&quot; : &nbsp; &nbsp; &nbsp; &nbsp; a -= 1 &nbsp; &nbsp; &nbsp; &nbsp; list.append(a) ans = 0 for j in list: &nbsp; &nbsp; if ans &lt; j: &nbsp; &nbsp; &nbsp; &nbsp; ans = j print (ans)#之所以不能直接 print(str(L).count(&quot;[&quot;))是因为列表里不是只有一条纵深线路，可能又好几个内嵌列表线路，但是有一个最深线路

最简单的方法

<pre class="prettyprint lang-py linenums">L = str(L)
record = 0
count =0
for i in range(0,len(L)):
 count += (L[i] == '[')
 count -= (L[i] == ']')
 record = max(record, count)
print record</pre>

dep = 0 deplist = [] for s in str(L): &nbsp;&nbsp;&nbsp; if&nbsp;&nbsp; s == '[': dep+=1 &nbsp;&nbsp;&nbsp; elif s == ']': dep-=1 &nbsp;&nbsp;&nbsp; deplist.append(dep) print max(deplist) 想一样去了

投机取巧的方法

maxdp=0 def deepit(l,n): &nbsp;&nbsp;&nbsp; global maxdp &nbsp;&nbsp;&nbsp; for e in l: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; if isinstance(e,list): &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; deepit(e,n+1) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; else: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; if n&gt;maxdp: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; maxdp=n deepit(L,0) print maxdp+1

注意全局变量声明glabal

print(str(L).count(']'))这样不行吗？

我和楼主一样，为什么不能通过？奇了怪

这题一定要绕吗？

s = str(L) print(s.count('[')) # 只找左边[ 的个数不就行了

s=str(L) left=0 mmax=0 for i in range(0,len(s)): &nbsp; &nbsp; if s[i]=='[': &nbsp; &nbsp; &nbsp; &nbsp; left+=1 &nbsp; &nbsp; elif s[i]==']': &nbsp; &nbsp; &nbsp; &nbsp; left-=1 &nbsp; &nbsp; mmax=max(mmax,left) print mmax

第150题：list深度

再谈数数II

最大整数

结尾0的个数III

计算GPA