2019-5-23

Author: JiayangWu

0114.二叉树展开为链表/0114-二叉树展开为链表.py

@@ -0,0 +1,29 @@
+# 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 flatten(self, root):
+        """
+        :type root: TreeNode
+        :rtype: None Do not return anything, modify root in-place instead.
+        """
+        
+        if not root or (not root.left and not root.right):
+            return root
+        
+        #先把左右子树捋直
+        self.flatten(root.left)
+        self.flatten(root.right)
+        
+        tmp = root.right #把捋直的右子树备份一下
+        
+        root.right = root.left #把捋直的左子树放到右边
+        root.left = None #记得把左子树置空
+        while(root.right): #找到现在右子树的最后一个node
+            root = root.right
+        root.right = tmp #把捋直的原来的右子树接上去
+        

0116.填充每个节点的下一个右侧节点指针/0116-填充每个节点的下一个右侧节点指针.py

@@ -0,0 +1,32 @@
+"""
+# Definition for a Node.
+class Node(object):
+    def __init__(self, val, left, right, next):
+        self.val = val
+        self.left = left
+        self.right = right
+        self.next = next
+"""
+class Solution(object):
+    def connect(self, root):
+        """
+        :type root: Node
+        :rtype: Node
+        """
+        #先排除掉不需要处理的情况
+        if not root or (not root.left and not root.right):
+            return root
+        
+        #某一个节点的左孩子的next一定是指向这个节点的右孩子
+        root.left.next = root.right
+        
+        #当某一个节点的next不为空的时候，这个节点的右孩子的next一定是指向该节点next的left
+        if root.next:
+            root.right.next = root.next.left
+        
+        #递归处理下一层
+        self.connect(root.left)
+        self.connect(root.right)
+  
+        return root
+                

0117.填充每个节点的下一个右侧节点指针II/0117-填充每个节点的下一个右侧节点指针II.py

@@ -0,0 +1,37 @@
+class Solution(object):
+    def connect(self, root):
+        """
+        :type root: Node
+        :rtype: Node
+        """
+        if not root or (not root.left and not root.right):
+            return root
+        
+        def findCousin(node, parent):
+            tmp = parent.next
+            while(tmp):
+                if tmp.left:
+                    node.next = tmp.left
+                    break
+                    
+                elif tmp.right:
+                    node.next = tmp.right
+                    break
+                tmp = tmp.next
+                
+        if root.left and root.right:
+            root.left.next = root.right
+            findCousin(root.right, root)
+            
+        elif root.left:
+            findCousin(root.left, root)
+            
+        elif root.right:
+            findCousin(root.right, root)
+        
+        # print root.val, root.right.next
+        self.connect(root.right)
+        self.connect(root.left)
+        
+                
+        return root

0139.单词拆分/0139-单词拆分.py

@@ -0,0 +1,17 @@
+class Solution(object):
+    def wordBreak(self, s, wordDict):
+        """
+        :type s: str
+        :type wordDict: List[str]
+        :rtype: bool
+        """
+        record = [0]#一开始从开头开始找
+        
+        for j in range(len(s) + 1):
+            for i in record:#在之前每一种找法的基础上找
+                if s[i : j] in wordDict: #找到一种可行的分法，说明最远可以拆分到j
+                    record.append(j)
+                    break
+        # print record
+        return record[-1] == len(s)
+                

0186.翻转字符串里的单词II/0186-翻转字符串里的单词II.py

@@ -0,0 +1,25 @@
+class Solution(object):
+    def reverseWords(self, string):
+        """
+        :type str: List[str]
+        :rtype: None Do not return anything, modify str in-place instead.
+        """
+        l = len(string)
+        if not l:
+            return 
+        
+        def reverse(start, end): #工具函数，功能是将string[start:end + 1]翻转
+            left, right = start, end
+            while(left < right):
+                string[left], string[right] = string[right], string[left]
+                left += 1
+                right -= 1
+        
+        reverse(0, l - 1) #先整个翻转
+        first_char_idx = 0
+        for i, x in enumerate(string):
+            if x == " ":
+                reverse(first_char_idx, i - 1) #再把每个单词进行翻转
+                first_char_idx = i + 1
+
+        reverse(first_char_idx, l - 1)#把最后一个单词翻转