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<h1 id="复制复杂链表"><a href="#复制复杂链表" class="headerlink" title="复制复杂链表"></a>复制复杂链表</h1><blockquote>
<p>题目：请实现 copyRandomList 函数，复制一个复杂链表。在复杂链表中，每个节点除了有一个 next 指针指向下一个节点，还有一个 random 指针指向链表中的任意节点或者 null。<br>这是去往<a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/fu-za-lian-biao-de-fu-zhi-lcof/">原题</a>的链接。</p>
</blockquote>
<h2 id="方法一：直接deepcopy"><a href="#方法一：直接deepcopy" class="headerlink" title="方法一：直接deepcopy"></a>方法一：直接deepcopy</h2><blockquote>
<p>复制分为浅拷贝和深拷贝。浅拷贝只复制指向某个对象的指针，而不复制对象本身，新旧对象还是共享同一块内存。但深拷贝会另外创造一个一模一样的对象，新对象跟原对象不共享内存，修改新对象不会改到原对象。</p>
</blockquote>
<p>明白了以上原理，对于 python 可直接调用相关函数：</p>
<pre><code>class Solution:
def copyRandomList(self, head: 'Node') -> 'Node':
return copy.deepcopy(head)</code></pre>
<h2 id="方法二：广度优先搜索-BFS-深度优先搜索（DFS）"><a href="#方法二：广度优先搜索-BFS-深度优先搜索（DFS）" class="headerlink" title="方法二：广度优先搜索 (BFS)/深度优先搜索（DFS）"></a>方法二：广度优先搜索 (BFS)/深度优先搜索（DFS）</h2><p>（先总结一下自己在做题过程中的困扰：初期学习算法，各种基础不够扎实，很多语句无法直接理解，我会尝试在colab中把大佬的代码跑一下，但是当涉及到调用class类中function时，经常会有问题，原因是题目会定义好一些内部class不可见，解决方法是直接在leetcode平常上跑代码即可。）</p>
<pre><code>"""
# Definition for a Node.
class Node:
def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None):
self.val = int(x)
self.next = next
self.random = random
"""
#这表示这个class node在内部被定义了，我们可以在soluton中直接调用
class Solution:
def copyRandomList(self, head: 'Node') -> 'Node':
if not head:
return None
dic = {} #set 作为hash表 用来存放node的value
cur = head
while cur:
dic[cur] = Node(cur.val)
cur = cur.next
for cur in dic:
if cur.next:
dic[cur].next = dic[cur.next]
if cur.random:
dic[cur].random = dic[cur.random]
return dic[head]</code></pre>
<blockquote>
<p>BFS 广度优先搜索，是从根结点开始，沿着树的宽度遍历树的节点，其应用主要是应用于走迷宫，从起点开始，找出到终点的最短路径。很多最短路径都基于此计算。这题是deep copy，我们利用hash表的特点，所谓哈希表（Hash table），是根据键（Key）而直接访问在内存存储位置的数据结构。</p>
</blockquote>
<p>这里的代码是搬运大佬的，并不是特别理解…<br>只好先来看一下方法三把…</p>
<h2 id="方法三：手写deepcopy"><a href="#方法三：手写deepcopy" class="headerlink" title="方法三：手写deepcopy"></a>方法三：手写deepcopy</h2><p>先来理解一下解题思路：</p>
<p>1.把链表每个节点复制一下（先不复制random），并紧跟被复制节点加入链表</p>
<p>2.给newNode的random赋值，如何赋值呢？可用node.next.random来表示newNode所要指向的random，用node.random.next来表示原来的node指向的random的复制品。令node.next.random = node.random.next</p>
<p>3.将newNode从链表中分离出来</p>
<p>python 3 代码：</p>
<pre><code>"""
# Definition for a Node.
class Node:
def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None):
self.val = int(x)
self.next = next
self.random = random
"""
class Solution:
def copyRandomList(self, head: 'Node') -> 'Node':
if not head: return
#1.把链表每个节点复制一下（先不复制random），并紧跟被节点加入链表
tmpt = head
while tmpt:
newnode = Node(tmpt.val,None,None)
#把原始node的值赋给newnode
newnode.next,tmpt.next = tmpt.next,newnode
#上面的语句可以用下面两个代替
#newnode.next = tmpt.next
#tmpt.next = newnode #把newnode放在临时变量tmpt.next中，把tmpt.next中的值赋给newnode.next
tmpt = tmpt.next.next
#将新的链表中新加入的节点的random赋值，可用node.next.random = node.random.next
tmpt = head
while tmpt:
if tmpt.random == None:
tmpt.next.random == None
else:
tmpt.next.random = tmpt.random.next
tmpt = tmpt.next.next
#将newNode从链表中分离出来
left = head
right = head.next
start = right
while left.next and right.next:
left.next = left.next.next
right.next = right.next.next
left = left.next
right = right.next
return start</code></pre>
<h1 id="最低票价问题"><a href="#最低票价问题" class="headerlink" title="最低票价问题"></a>最低票价问题</h1><blockquote>
<p>在一个火车旅行很受欢迎的国度，你提前一年计划了一些火车旅行。在接下来的一年里，你要旅行的日子将以一个名为 days 的数组给出。每一项是一个从 1 到 365 的整数。<br>火车票有三种不同的销售方式：<br>一张为期一天的通行证售价为 costs[0] 美元；<br>一张为期七天的通行证售价为 costs[1] 美元；<br>一张为期三十天的通行证售价为 costs[2] 美元。<br>通行证允许数天无限制的旅行。 例如，如果我们在第 2 天获得一张为期 7 天的通行证，那么我们可以连着旅行 7 天：第 2 天、第 3 天、第 4 天、第 5 天、第 6 天、第 7 天和第 8 天。<br>返回你想要完成在给定的列表 days 中列出的每一天的旅行所需要的最低消费。<br>这是去往<a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/minimum-cost-for-tickets/">原题</a>的链接。</p>
</blockquote>
<p>题目分析：本题考察的是一维dynamic programming，解题思路是新建dp array用来存放每天的票价，通过判断dp[i]中的index i是否与days array中的days[index]相对应，来建立两个array之间的联系。难点在于如果i!=days[index]的话，dp[i]=dp[i-1]。如果=的话，那么就需要从三种方案dp[max(0,i-1)]+costs[0],dp[max(0,i-7)]+costs[1],dp[max(0,i-30)]+costs[2]中选出最小的，从而作为iretation的内部，最后取dp[-1]，则为最后花的最少的票价。</p>
<p>代码如下：</p>
<pre><code>class Solution:
def mincostTickets(self, days: List[int], costs: List[int]) -> int:
#对dp array进行初始化,dp array的长度应该是days最后一个元素➕1
dp=[0 for _ in range(days[-1]+1)]
#新增一个变量days_index,标记应该处理 days 数组中哪一个元素,并初始化为0
#这样可以避免多增加一个指针j，指向days[j],从而降低时间复杂度
days_index=0
for i in range(1,len(dp)):
if i!= days[days_index]:
#若当前天数不是待处理天数，则其花费费用和前一天相同
dp[i]=dp[i-1]
else:
#若 i 走到了待处理天数，则从三种方式中选一个最小的
dp[i]=min(dp[max(0,i-1)]+costs[0],
dp[max(0,i-7)]+costs[1],
dp[max(0,i-30)]+costs[2])
days_index += 1
return dp[-1] #返回最后一天对应的费用</code></pre>
<h1 id="棋盘问题（二维Dynamic-Programming）"><a href="#棋盘问题（二维Dynamic-Programming）" class="headerlink" title="棋盘问题（二维Dynamic Programming）"></a>棋盘问题（二维Dynamic Programming）</h1><blockquote>
<p>给你一个 m * n 的网格，其中每个单元格不是 0（空）就是 1（障碍物）。每一步，您都可以在空白单元格中上、下、左、右移动。<br>如果您 最多 可以消除 k 个障碍物，请找出从左上角 (0, 0) 到右下角 (m-1, n-1) 的最短路径，并返回通过该路径所需的步数。如果找不到这样的路径，则返回 -1。<br>这是去往<a target="_blank" rel="noopener" href="https://leetcode-cn.com/problems/shortest-path-in-a-grid-with-obstacles-elimination/">原题</a>的链接。</p>
</blockquote>
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