lintcode 刷题记录··

单例模式，用C#实现过单例模式，python区别就是类里边的静态方法写法不一样，python叫类方法,函数之前加@classmethod

class Solution:
    # @return: The same instance of this class every time
    __m_Solution=None
    
    @classmethod
    def getInstance(self):
        # write your code here
        if self.__m_Solution == None:
            self.__m_Solution=Solution()
        return self.__m_Solution

    def __init__(self):
        pass

 

 

遍历二叉树路径，输出与给定值相同的所有路径，这里是递归调用的，本来像把递归改成循环的 但是高了好久 没搞出来····

class Solution：
    # @param {int} target an integer
    # @return {int[][]} all valid paths
    def binaryTreePathSum(self, root, target):
        # Write your code here
        res = []
        path = []
        import copy
        def getallpath(current, path, res):
            if current == None: return
            path.append(current.val)
            if current.left != None:
                getallpath(current.left,copy.deepcopy(path), res)
            if current.right != None:
                getallpath(current.right,copy.deepcopy(path),res)
            if current.left is None and current.right is None:
                res.append(path)

        getallpath(root,path,res)
        ps=[]
        for p in res:
            sum=0
            for v in p:
                sum=sum+v
            if sum==target:ps.append(p)
        return  ps
        

 三角形计数 给定一个整数数组，在该数组中，寻找三个数，分别代表三角形三条边的长度，问，可以寻找到多少组这样的三个数来组成三角形

直接遍历会提示超过时间限制

选择排序加二分查找的方法来提高效率，python默认的递归有上线，需要设置，  还有二分查找的方法是先确认最大最小边然后找第三边，这样条件就变成一个

class Solution:
    # @param S: a list of integers
    # @return: a integer
    def triangleCount(self, S):
        # write your code here
        def quicksort(a):  # 快速排序
            if len(a) < 2:
                return a
            else:
                pivot = a[0]
                less = [i for i in a[1:] if i <= pivot]
                greator = [i for i in a[1:] if i > pivot]
                return quicksort(less) + [pivot] + quicksort(greator)

        n = len(S)
        if n < 3: return
        import sys
        sys.setrecursionlimit(1000000)
        S = quicksort(S)
        sum = 0
        for i in range(n):
            for j in range(i + 1, n):
                l = i + 1
                r = j
                target = S[j] - S[i]
                while l < r:
                    mid = (l + r) // 2
                    if S[mid] > target:
                        r = mid
                    else:
                        l = mid + 1
                sum = sum + j - l

        return sum

 将一个二叉查找树按照中序遍历转换成双向链表。

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        this.val = val
        this.left, this.right = None, None
Definition of Doubly-ListNode
class DoublyListNode(object):

    def __init__(self, val, next=None):
        self.val = val
        self.next = self.prev = next
"""

class Solution:
    """
    @param root, the root of tree
    @return: a doubly list node
    """
    def bstToDoublyList(self, root):
        # Write your code here
        l=[]
        if root is None:return
        stack=[]
        current=root
        dlist=None
        cur=None
        last=None
        while len(stack)> 0 or current is not None:
            while current is not None:
                stack.append(current)
                current=current.left
            current=stack.pop()
            l.append(current.val)
            cur=DoublyListNode(current.val)
            cur.prev=last
            if last is not None: last.next=cur
            last=cur
            if dlist is None :dlist=cur
            current=current.right
        return dlist