first commit

Author: princewen

100_same_tree.py

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+"""
+
+Given two binary trees, write a function to check if they are equal or not.
+
+Two binary trees are considered equal if they are structurally identical and the nodes have the same value.
+
+"""
+
+"""my solution
+采用中序遍历的方式
+"""
+
+# 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 isSameTree(self, p, q):
+        """
+        :type p: TreeNode
+        :type q: TreeNode
+        :rtype: bool
+        """
+
+        if p != None and q != None:
+            if self.isSameTree(p.left, q.left) and p.val == q.val and self.isSameTree(p.right, q.right):
+                return True
+            else:
+                return False
+        else:
+            if p == None and q == None:
+                return True
+            else:
+                return False
+
+
+"""more simple way"""
+def isSameTree(self, p, q):
+    if p and q:
+        return p.val == q.val and self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)
+    return p is q

14_longgest_common_prefix.py

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+"""
+
+Write a function to find the longest common prefix string amongst an array of strings.
+
+"""
+
+
+class Solution(object):
+    def longestCommonPrefix(self, strs):
+        """
+        :type strs: List[str]
+        :rtype: str
+        """
+        if len(strs) == 0:
+            return ''
+        cpre = strs[0]
+        for onestr in strs[1:]:
+            i = 0
+            while i < len(cpre) and i < len(onestr) and cpre[i] == onestr[i]:
+                i = i + 1
+            cpre = cpre[:i]
+
+        return cpre
+
+
+"""an answer using zip"""
+
+
+def longestCommonPrefix(self, strs):
+    """
+    :type strs: List[str]
+    :rtype: str
+    """
+    if not strs: return ''
+    l = 0
+    for cg in zip(*strs):
+        if len(set(cg)) > 1:
+            return strs[0][:l]
+        l += 1
+    return strs[0][:l]

1_two_sum.py

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+"""
+
+Given an array of integers, return indices of the two numbers such that they add up to a specific target.
+
+You may assume that each input would have exactly one solution, and you may not use the same element twice.
+
+Example:
+Given nums = [2, 7, 11, 15], target = 9,
+
+Because nums[0] + nums[1] = 2 + 7 = 9,
+return [0, 1].
+
+"""
+
+"""my solution
+首先创建一个字典，用于保存已经遍历过的元素以及他们在字典中的位置
+如果差值在字典中，则返回字典中的索引值以及该元素的索引值
+如果差值不再字典中，将该元素加入字典
+"""
+
+
+class Solution:
+    def twoSum(self, nums, target):
+        """
+        :type nums: List[int]
+        :type target: int
+        :rtype: List[int]
+        """
+        num_dict = dict()
+        for index, value in enumerate(nums):
+            sub = target - value
+            if sub in num_dict.keys():
+                return [num_dict[sub], index]
+            else:
+                num_dict[value] = index
+
+        return []
+
+
+"""标准答案1
+跟我的思路差不多，是把差值存入字典中
+而且考虑了nums长度小于1的情况
+"""
+
+
+def twoSum(self, nums, target):
+    if len(nums) <= 1:
+        return False
+    buff_dict = {}
+    for i in range(len(nums)):
+        if nums[i] in buff_dict:
+            return [buff_dict[nums[i]], i]
+        else:
+            buff_dict[target - nums[i]] = i

20_Valid_Parentheses.py

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+"""
+
+Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
+
+The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.
+
+"""
+
+"""my solution
+使用一个栈，如果是前半部分就进栈，如果是后半部分就出栈 ，并测试是否匹配
+"""
+class Solution(object):
+    def isValid(self, s):
+        """
+        :type s: str
+        :rtype: bool
+        """
+        if len(s) % 2 != 0:
+            return False
+        stack = []
+        push_list = ['(','[','{']
+        pop_list = [')',']','}']
+        for single_char in s:
+            if single_char in push_list:
+                stack.append(single_char)
+            if single_char in pop_list:
+                if single_char == ')':
+                    if len(stack)==0 or stack.pop() != '(':
+                        return False
+                elif single_char ==']':
+                    if len(stack)==0 or stack.pop() != '[':
+                        return False
+                elif single_char == '}':
+                    if len(stack)==0 or stack.pop() != '{':
+                        return False
+        if len(stack) == 0:
+            return True
+        else:
+            return False
+
+"""前面我定义了两个list，其实可以定义成一个dict 最后的判断stack是否为空也可以写成一条语句"""
+class Solution:
+    # @return a boolean
+    def isValid(self, s):
+        stack = []
+        dict = {"]":"[", "}":"{", ")":"("}
+        for char in s:
+            if char in dict.values():
+                stack.append(char)
+            elif char in dict.keys():
+                if stack == [] or dict[char] != stack.pop():
+                    return False
+            else:
+                return False
+        return stack == []

21_Merge_Two_Sorted_Lists.py

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+"""
+
+Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
+
+"""
+
+
+# Definition for singly-linked list.
+class ListNode(object):
+    def __init__(self, x):
+        self.val = x
+        self.next = None
+
+class Solution(object):
+    def mergeTwoLists(self, l1, l2):
+        """
+        :type l1: ListNode
+        :type l2: ListNode
+        :rtype: ListNode
+        """
+        head = cur = ListNode(0)
+        while l1 and l2:
+            if l1.val > l2.val:
+                cur.next = l2
+                l2 = l2.next
+            else:
+                cur.next = l1
+                l1 = l1.next
+            cur = cur.next
+        """or 返回第一个为真的值"""
+        cur.next = l1 or l2
+        return head.next

26_Remove_Duplicates_from_Sorted_Array.py

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+"""
+
+Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.
+
+Do not allocate extra space for another array, you must do this in place with constant memory.
+
+For example,
+Given input array nums = [1,1,2],
+
+Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn't matter what you leave beyond the new length.
+
+"""
+"""my solution
+这里要注意一个额外的细节，那就是要求把不同的元素放到数组的前几位，不能简单的返回一个count值
+"""
+
+
+class Solution(object):
+    def removeDuplicates(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        if nums == []:
+            return 0
+        count = 1
+        cur = nums[0]
+        for i in nums:
+            if i == cur:
+                continue
+            else:
+                cur = i
+                nums[count] = i
+                count += 1
+
+        return count
+
+
+"""
+other solutions
+"""
+
+class Solution:
+    # @param a list of integers
+    # @return an integer
+    def removeDuplicates(self, A):
+        if not A:
+            return 0
+
+        newTail = 0
+
+        for i in range(1, len(A)):
+            if A[i] != A[newTail]:
+                newTail += 1
+                A[newTail] = A[i]
+
+        return newTail + 1

27_Remove_Element.py

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+"""
+
+Given an array and a value, remove all instances of that value in place and return the new length.
+
+Do not allocate extra space for another array, you must do this in place with constant memory.
+
+The order of elements can be changed. It doesn't matter what you leave beyond the new length.
+
+Example:
+Given input array nums = [3,2,2,3], val = 3
+
+Your function should return length = 2, with the first two elements of nums being 2.
+
+
+
+"""
+
+""" my solution 
+借鉴26题的思路，直接遍历，由于前面的已经遍历过，可以直接替换屌元素的值
+"""
+
+
+class Solution(object):
+    def removeElement(self, nums, val):
+        """
+        :type nums: List[int]
+        :type val: int
+        :rtype: int
+        """
+        index = 0
+        for i in range(0, len(nums)):
+            if nums[i] != val:
+                nums[index] = nums[i]
+                index = index + 1
+
+        return index
+
+
+"""other solution
+交换的思路
+"""