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CyC2018 2018-06-30 14:11:16 +08:00
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2
notes/Java 虚拟机.md
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@ -261,7 +261,7 @@ obj = null;
finalize() 类似 C++ 的析构函数，用来做关闭外部资源等工作。但是 try-finally 等方式可以做的更好，并且该方法运行代价高昂，不确定性大，无法保证各个对象的调用顺序，因此最好不要使用。
当一个对象可被回收时，如果需要执行该对象的 finalize() 方法，那么就有可能通过在该方法中让对象重新被引用，从而实现自救。
当一个对象可被回收时，如果需要执行该对象的 finalize() 方法，那么就有可能通过在该方法中让对象重新被引用，从而实现自救。自救只能进行一次，如果回收的对象之前调用了 finalize() 方法自救，后面回收时不会调用 finalize() 方法。
## 垃圾收集算法

2
notes/Linux.md
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@ -130,6 +130,8 @@ info 与 man 类似，但是 info 将文档分成一个个页面，每个页面
/usr/local/bin:/usr/bin:/usr/local/sbin:/usr/sbin:/home/dmtsai/.local/bin:/home/dmtsai/bin
```
env 命令可以获取当前终端的环境变量。
## sudo
sudo 允许一般用户使用 root 可执行的命令，不过只有在 /etc/sudoers 配置文件中添加的用户才能使用该指令。

164
notes/剑指 offer 题解.md
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@ -1062,20 +1062,21 @@ public void reOrderArray(int[] nums) {
<div align="center"> <img src="../pics//207c1801-2335-4b1b-b65c-126a0ba966cb.png" width="500"/> </div><br>
```java
public ListNode FindKthToTail(ListNode head, int k) {
public ListNode FindKthToTail(ListNode head, int k)
{
if (head == null)
return null;
ListNode fast, slow;
fast = slow = head;
while (fast != null && k-- > 0)
fast = fast.next;
ListNode P1 = head;
while (P1 != null && k-- > 0)
P1 = P1.next;
if (k > 0)
return null;
while (fast != null) {
fast = fast.next;
slow = slow.next;
ListNode P2 = head;
while (P1 != null) {
P1 = P1.next;
P2 = P2.next;
}
return slow;
return P2;
}
```
@ -1083,6 +1084,12 @@ public ListNode FindKthToTail(ListNode head, int k) {
[NowCoder](https://www.nowcoder.com/practice/253d2c59ec3e4bc68da16833f79a38e4?tpId=13&tqId=11208&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
## 题目描述
一个链表中包含环，请找出该链表的环的入口结点。
要求不能使用额外的空间。
## 解题思路
使用双指针，一个指针 fast 每次移动两个节点，一个指针 slow 每次移动一个节点。因为存在环，所以两个指针必定相遇在环中的某个节点上。假设相遇点在下图的 z1 位置，此时 fast 移动的节点数为 x+2y+z，slow 为 x+y，由于 fast 速度比 slow 快一倍，因此 x+2y+z=2(x+y)，得到 x=z。
@ -1092,16 +1099,15 @@ public ListNode FindKthToTail(ListNode head, int k) {
<div align="center"> <img src="../pics//71363383-2d06-4c63-8b72-c01c2186707d.png" width="600"/> </div><br>
```java
public ListNode EntryNodeOfLoop(ListNode pHead) {
public ListNode EntryNodeOfLoop(ListNode pHead)
{
if (pHead == null || pHead.next == null)
return null;
ListNode slow = pHead, fast = pHead;
while (fast != null && fast.next != null) {
do {
fast = fast.next.next;
slow = slow.next;
if (slow == fast)
break;
}
} while (slow != fast);
fast = pHead;
while (slow != fast) {
slow = slow.next;
@ -1159,7 +1165,8 @@ public ListNode ReverseList(ListNode head) {
### 递归
```java
public ListNode Merge(ListNode list1, ListNode list2) {
public ListNode Merge(ListNode list1, ListNode list2)
{
if (list1 == null)
return list2;
if (list2 == null)
@ -1177,7 +1184,8 @@ public ListNode Merge(ListNode list1, ListNode list2) {
### 迭代
```java
public ListNode Merge(ListNode list1, ListNode list2) {
public ListNode Merge(ListNode list1, ListNode list2)
{
ListNode head = new ListNode(-1);
ListNode cur = head;
while (list1 != null && list2 != null) {
@ -1209,20 +1217,22 @@ public ListNode Merge(ListNode list1, ListNode list2) {
## 解题思路
```java
public boolean HasSubtree(TreeNode root1, TreeNode root2) {
public boolean HasSubtree(TreeNode root1, TreeNode root2)
{
if (root1 == null || root2 == null)
return false;
return isSubtree(root1, root2) || HasSubtree(root1.left, root2) || HasSubtree(root1.right, root2);
return isSubtreeWithRoot(root1, root2) || HasSubtree(root1.left, root2) || HasSubtree(root1.right, root2);
}
private boolean isSubtree(TreeNode root1, TreeNode root2) {
private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2)
{
if (root2 == null)
return true;
if (root1 == null)
return false;
if (root1.val != root2.val)
return false;
return isSubtree(root1.left, root2.left) && isSubtree(root1.right, root2.right);
return isSubtreeWithRoot(root1.left, root2.left) && isSubtreeWithRoot(root1.right, root2.right);
}
```
@ -1237,7 +1247,8 @@ private boolean isSubtree(TreeNode root1, TreeNode root2) {
## 解题思路
```java
public void Mirror(TreeNode root) {
public void Mirror(TreeNode root)
{
if (root == null)
return;
swap(root);
@ -1245,7 +1256,8 @@ public void Mirror(TreeNode root) {
Mirror(root.right);
}
private void swap(TreeNode root) {
private void swap(TreeNode root)
{
TreeNode t = root.left;
root.left = root.right;
root.right = t;
@ -1263,13 +1275,15 @@ private void swap(TreeNode root) {
## 解题思路
```java
boolean isSymmetrical(TreeNode pRoot) {
boolean isSymmetrical(TreeNode pRoot)
{
if (pRoot == null)
return true;
return isSymmetrical(pRoot.left, pRoot.right);
}
boolean isSymmetrical(TreeNode t1, TreeNode t2) {
boolean isSymmetrical(TreeNode t1, TreeNode t2)
{
if (t1 == null && t2 == null)
return true;
if (t1 == null || t2 == null)
@ -1293,7 +1307,8 @@ boolean isSymmetrical(TreeNode t1, TreeNode t2) {
## 解题思路
```java
public ArrayList<Integer> printMatrix(int[][] matrix) {
public ArrayList<Integer> printMatrix(int[][] matrix)
{
ArrayList<Integer> ret = new ArrayList<>();
int r1 = 0, r2 = matrix.length - 1, c1 = 0, c2 = matrix[0].length - 1;
while (r1 <= r2 && c1 <= c2) {
@ -1324,24 +1339,28 @@ public ArrayList<Integer> printMatrix(int[][] matrix) {
## 解题思路
```java
private Stack<Integer> stack = new Stack<>();
private Stack<Integer> dataStack = new Stack<>();
private Stack<Integer> minStack = new Stack<>();
public void push(int node) {
stack.push(node);
public void push(int node)
{
dataStack.push(node);
minStack.push(minStack.isEmpty() ? node : Math.min(minStack.peek(), node));
}
public void pop() {
stack.pop();
public void pop()
{
dataStack.pop();
minStack.pop();
}
public int top() {
return stack.peek();
public int top()
{
return dataStack.peek();
}
public int min() {
public int min()
{
return minStack.peek();
}
```
@ -1359,12 +1378,13 @@ public int min() {
使用一个栈来模拟压入弹出操作。
```java
public boolean IsPopOrder(int[] pushA, int[] popA) {
int n = pushA.length;
public boolean IsPopOrder(int[] pushSequence, int[] popSequence)
{
int n = pushSequence.length;
Stack<Integer> stack = new Stack<>();
for (int pushIndex = 0, popIndex = 0; pushIndex < n; pushIndex++) {
stack.push(pushA[pushIndex]);
while (popIndex < n && stack.peek() == popA[popIndex]) {
stack.push(pushSequence[pushIndex]);
while (popIndex < n && stack.peek() == popSequence[popIndex]) {
stack.pop();
popIndex++;
}
@ -1392,21 +1412,20 @@ public boolean IsPopOrder(int[] pushA, int[] popA) {
不需要使用两个队列分别存储当前层的节点和下一层的节点，因为在开始遍历一层的节点时，当前队列中的节点数就是当前层的节点数，只要控制遍历这么多节点数，就能保证这次遍历的都是当前层的节点。
```java
public ArrayList<Integer> PrintFromTopToBottom(TreeNode root) {
public ArrayList<Integer> PrintFromTopToBottom(TreeNode root)
{
Queue<TreeNode> queue = new LinkedList<>();
ArrayList<Integer> ret = new ArrayList<>();
if (root == null)
return ret;
queue.add(root);
while (!queue.isEmpty()) {
int cnt = queue.size();
while (cnt-- > 0) {
TreeNode t = queue.poll();
if (t.left != null)
queue.add(t.left);
if (t.right != null)
queue.add(t.right);
if (t == null)
continue;
ret.add(t.val);
queue.add(t.left);
queue.add(t.right);
}
}
return ret;
@ -1424,10 +1443,9 @@ public ArrayList<Integer> PrintFromTopToBottom(TreeNode root) {
## 解题思路
```java
ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {
ArrayList<ArrayList<Integer>> Print(TreeNode pRoot)
{
ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
if (pRoot == null)
return ret;
Queue<TreeNode> queue = new LinkedList<>();
queue.add(pRoot);
while (!queue.isEmpty()) {
@ -1435,12 +1453,13 @@ ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {
int cnt = queue.size();
while (cnt-- > 0) {
TreeNode node = queue.poll();
if (node == null)
continue;
list.add(node.val);
if (node.left != null)
queue.add(node.left);
if (node.right != null)
queue.add(node.right);
}
if (list.size() != 0)
ret.add(list);
}
return ret;
@ -1458,10 +1477,9 @@ ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {
## 解题思路
```java
public ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {
public ArrayList<ArrayList<Integer>> Print(TreeNode pRoot)
{
ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
if (pRoot == null)
return ret;
Queue<TreeNode> queue = new LinkedList<>();
queue.add(pRoot);
boolean reverse = false;
@ -1470,15 +1488,16 @@ public ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {
int cnt = queue.size();
while (cnt-- > 0) {
TreeNode node = queue.poll();
if (node == null)
continue;
list.add(node.val);
if (node.left != null)
queue.add(node.left);
if (node.right != null)
queue.add(node.right);
}
if (reverse)
Collections.reverse(list);
reverse = !reverse;
if (list.size() != 0)
ret.add(list);
}
return ret;
@ -1500,13 +1519,15 @@ public ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {
## 解题思路
```java
public boolean VerifySquenceOfBST(int[] sequence) {
public boolean VerifySquenceOfBST(int[] sequence)
{
if (sequence == null || sequence.length == 0)
return false;
return verify(sequence, 0, sequence.length - 1);
}
private boolean verify(int[] sequence, int first, int last) {
private boolean verify(int[] sequence, int first, int last)
{
if (last - first <= 1)
return true;
int rootVal = sequence[last];
@ -1537,18 +1558,20 @@ private boolean verify(int[] sequence, int first, int last) {
```java
private ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
public ArrayList<ArrayList<Integer>> FindPath(TreeNode root, int target) {
public ArrayList<ArrayList<Integer>> FindPath(TreeNode root, int target)
{
backtracking(root, target, new ArrayList<>());
return ret;
}
private void backtracking(TreeNode node, int target, ArrayList<Integer> path) {
private void backtracking(TreeNode node, int target, ArrayList<Integer> path)
{
if (node == null)
return;
path.add(node.val);
target -= node.val;
if (target == 0 && node.left == null && node.right == null) {
ret.add(new ArrayList(path));
ret.add(new ArrayList<>(path));
} else {
backtracking(node.left, target, path);
backtracking(node.right, target, path);
@ -1582,7 +1605,8 @@ private void backtracking(TreeNode node, int target, ArrayList<Integer> path) {
<div align="center"> <img src="../pics//8f3b9519-d705-48fe-87ad-2e4052fc81d2.png" width="600"/> </div><br>
```java
public RandomListNode Clone(RandomListNode pHead) {
public RandomListNode Clone(RandomListNode pHead)
{
if (pHead == null)
return null;
// 插入新节点
@ -1629,14 +1653,14 @@ public RandomListNode Clone(RandomListNode pHead) {
private TreeNode pre = null;
private TreeNode head = null;
public TreeNode Convert(TreeNode root) {
if (root == null)
return null;
public TreeNode Convert(TreeNode root)
{
inOrder(root);
return head;
}
private void inOrder(TreeNode node) {
private void inOrder(TreeNode node)
{
if (node == null)
return;
inOrder(node.left);
@ -1661,22 +1685,23 @@ private void inOrder(TreeNode node) {
## 解题思路
```java
public class Solution {
private String deserializeStr;
public String Serialize(TreeNode root) {
public String Serialize(TreeNode root)
{
if (root == null)
return "#";
return root.val + " " + Serialize(root.left) + " " + Serialize(root.right);
}
public TreeNode Deserialize(String str) {
public TreeNode Deserialize(String str)
{
deserializeStr = str;
return Deserialize();
}
private TreeNode Deserialize() {
private TreeNode Deserialize()
{
if (deserializeStr.length() == 0)
return null;
int index = deserializeStr.indexOf(" ");
@ -1690,7 +1715,6 @@ public class Solution {
t.right = Deserialize();
return t;
}
}
```
# 38. 字符串的排列