pythonpitfall-创新互联

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#对于迭代器，产生的是对象临时副本，不能修改对象。
for i in lists:
  change(i)

#在循环中修改迭代器迭代对象的地址，需要在while循环中重新迭代，不能继续在原有循环中

while list:
　  next=[]
for i in list:
      do some thing
      next.append(i)
   list=next

is 使用id 判断，==使用value判断，少用is

http://www.toptal.com/python#hiring-guide pythonpitfall

range(x)，ｘ不能取到，写代码时一定记住,range(len(s))刚好可以，因为下标减一

Python 负数取模运算

http://stackoverflow.com/questions/12946116/twos-complement-binary-in-python


https://leetcode.com/problems/single-number-ii/

class Solution:
# @param A, a list of integer
# @return an integerdef singleNumber(self, A):
    ans= 0
for i in xrange(0,32):
        count= 0
for a in A:
if ((a >> i) & 1):
                count+=1
        ans|= ((count%3) << i)
return self.convert(ans)

def convert(self,x):
if x >= 2**31:
        x-= 2**32
  return x

https://www.hackerrank.com/challenges/flipping-bits/editorial
python 按位取反　使用string replace
检查32位是否溢出

if res>= (1<<32)-1:
    return 0


//python 在函数参数中传递引用时,函数中的修改不会改变这个引用在函数作用域以外的值.

https://leetcode.com/problems/convert-sorted-list-to-binary-search-tree/

class Solution {
public:
    TreeNode*sortedListToBST(ListNode *head) {
int count = 0;
for (auto cur = head; cur;) {
    cur= cur->next;
++count;
}
return helper(head, count);
}
private:
TreeNode*helper(ListNode *&p, int n) {
if (!n) return NULL;
auto left= helper(p, n / 2);
auto res= new TreeNode(p->val);
p= p->next;
res->left = left;
res->right = helper(p, n - n / 2 - 1);
return res;
}
};

# not right, p changed in c++ while head not changed in pythonclass Solution:
def f(self,head,num):
if num<=0:
return None
        left=self.f(head,num/2)
        res=TreeNode(head.val)
        head=head.next
        res.left,res.right=left,self.f(head,num-num/2-1)
return res

# @param {ListNode} head  # @return {TreeNode}  def sortedListToBST(self, head):
        tmp,num=head,0
while tmp:
            tmp=tmp.next
            num+=1

   return self.f(head,num)

#can be solved with a global variable
class Solution:
def f(self,l,r):
if l>r:
return None
        m=(l+r)/2

        left=self.f(l,m-1)
        a=TreeNode(self.start.val)
        self.start=self.start.next
        right=self.f(m+1,r)

        a.left,a.right=left,right
return a

# @param {ListNode} head  # @return {TreeNode}  def sortedListToBST(self, head):
if head is None:
return None
        self.start=head

        tmp,count=head,0
while tmp:
            count+=1
            tmp=tmp.next
return self.f(0,count-1)

enumerate, start means start from the wanted idx
a = [1, 2, 3]

>>> for idx, item in enumerate(a, start=2):
...     print idx, item
... 
2 13 24 3

better to do this
>>> for idx, item in enumerate(a[1:], start=2):
...    print idx, item
...
2 2
3 3

  
maxlen= reduce(max, map(len, dict), 0)

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pythonpitfall-创新互联

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