Python - LeetCode使用到的好用Fun.

發表於

1. List Init

1.var = [x for x in x]：快速定義List的使用方式，可以搭配if來使用
2.range(start:end)：產生range的int
3.len() : 長度

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checkMap = [[0 for y in range(nYCount)] for x in range(nXCount)]
nLength = len(checkMap)

基本LeetCode

2. Object Init

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#1.list() : LIST         []
#2.set() : 獨立不重覆    {}
#3.dict() : 字典         {{}}
#4.tuple() : 不可改變   ()

	listVar = list([5,3,3])
	setVar = set([5,3,3])
	
	print(listVar)
	print(setVar)
	
	#[5, 3, 3]
	#{3, 5}

基本LeetCode 771 807 595 804 491 832 728

3. String

1.count : 計算內容物的次數

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class Solution:
   def judgeCircle(self, moves):
       return moves.count('L') == moves.count('R') and moves.count('U') == moves.count('D') 
       """
       :type moves: str
       :rtype: bool
       """

基本LeetCode 657

4. List[int]

1.index : 找到對應的index
2.max : 取出最大值

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return (A.index(max(A)))

基本LeetCode 852

5. binary Tree

binary Tree的使用範例

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# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def mergeTrees(self, t1, t2):
        NewNode = TreeNode(0)
        def nextLoop(tCurrent,tcopy):
            if tcopy == None:
                return
            tCurrent.val += tcopy.val
            print(tcopy.val)
            if tcopy.left is not None:
                if tCurrent.left == None:
                    tCurrent.left = TreeNode(0)
                nextLoop(tCurrent.left,tcopy.left)
            if tcopy.right is not None:
                if tCurrent.right == None:
                    tCurrent.right = TreeNode(0)
                nextLoop(tCurrent.right,tcopy.right)
        nextLoop(NewNode,t1)
        nextLoop(NewNode,t2)
        if t1 == None and t2 == None:
            NewNode = None
        return NewNode
        """
        :type t1: TreeNode
        :type t2: TreeNode
        :rtype: TreeNode
        """

基本LeetCode 617