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‎problems/0018.四数之和.md‎

Lines changed: 4 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -35,7 +35,7 @@
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四数之和,和[15.三数之和](https://programmercarl.com/0015.三数之和.html)是一个思路,都是使用双指针法, 基本解法就是在[15.三数之和](https://programmercarl.com/0015.三数之和.html) 的基础上再套一层for循环。
3737

38-
但是有一些细节需要注意,例如: 不要判断`nums[k] > target` 就返回了,三数之和 可以通过 `nums[i] > 0` 就返回了,因为 0 已经是确定的数了,四数之和这道题目 target是任意值。比如:数组是`[-4, -3, -2, -1]`,`target`是`-10`,不能因为`-4 > -10`而跳过。但是我们依旧可以去做剪枝,逻辑变成`nums[i] > target && (nums[i] >=0 || target >= 0)`就可以了。
38+
但是有一些细节需要注意,例如: 不要判断`nums[k] > target` 就返回了,三数之和 可以通过 `nums[i] > 0` 就返回了,因为 0 已经是确定的数了,四数之和这道题目 target是任意值。比如:数组是`[-4, -3, -2, -1]`,`target`是`-10`,不能因为`-4 > -10`而跳过。但是我们依旧可以去做剪枝,逻辑变成`nums[k] > target && (nums[k] >=0 || target >= 0)`就可以了。
3939

4040
[15.三数之和](https://programmercarl.com/0015.三数之和.html)的双指针解法是一层for循环num[i]为确定值,然后循环内有left和right下标作为双指针,找到nums[i] + nums[left] + nums[right] == 0。
4141

@@ -253,7 +253,7 @@ public class Solution {
253253
for (int k = 0; k < nums.length; k++) {
254254
// 剪枝处理
255255
if (nums[k] > target && nums[k] >= 0) {
256-
break;
256+
break; // 此处的break可以等价于return result;
257257
}
258258
// 对nums[k]去重
259259
if (k > 0 && nums[k] == nums[k - 1]) {
@@ -262,7 +262,7 @@ public class Solution {
262262
for (int i = k + 1; i < nums.length; i++) {
263263
// 第二级剪枝
264264
if (nums[k] + nums[i] > target && nums[k] + nums[i] >= 0) {
265-
break;
265+
break; // 注意是break到上一级for循环,如果直接return result;会有遗漏
266266
}
267267
// 对nums[i]去重
268268
if (i > k + 1 && nums[i] == nums[i - 1]) {
@@ -802,3 +802,4 @@ end
802802
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
803803
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
804804
</a>
805+

‎problems/0019.删除链表的倒数第N个节点.md‎

Lines changed: 30 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -129,6 +129,36 @@ class Solution {
129129
}
130130
```
131131

132+
133+
```java
134+
class Solution {
135+
public ListNode removeNthFromEnd(ListNode head, int n) {
136+
// 创建一个新的哑节点,指向原链表头
137+
ListNode s = new ListNode(-1, head);
138+
// 递归调用remove方法,从哑节点开始进行删除操作
139+
remove(s, n);
140+
// 返回新链表的头(去掉可能的哑节点)
141+
return s.next;
142+
}
143+
144+
public int remove(ListNode p, int n) {
145+
// 递归结束条件:如果当前节点为空,返回0
146+
if (p == null) {
147+
return 0;
148+
}
149+
// 递归深入到下一个节点
150+
int net = remove(p.next, n);
151+
// 如果当前节点是倒数第n个节点,进行删除操作
152+
if (net == n) {
153+
p.next = p.next.next;
154+
}
155+
// 返回当前节点的总深度
156+
return net + 1;
157+
}
158+
}
159+
```
160+
161+
132162
### Python:
133163

134164
```python

‎problems/0020.有效的括号.md‎

Lines changed: 25 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -166,12 +166,35 @@ class Solution {
166166
deque.pop();
167167
}
168168
}
169-
//最后判断栈中元素是否匹配
169+
//遍历结束,如果栈为空,则括号全部匹配
170170
return deque.isEmpty();
171171
}
172172
}
173173
```
174174

175+
```java
176+
// 解法二
177+
// 对应的另一半一定在栈顶
178+
class Solution {
179+
public boolean isValid(String s) {
180+
Stack<Character> stack = new Stack<>();
181+
for(char c : s.toCharArray()){
182+
// 有对应的另一半就直接消消乐
183+
if(c == ')' && !stack.isEmpty() && stack.peek() == '(')
184+
stack.pop();
185+
else if(c == '}' && !stack.isEmpty() && stack.peek() == '{')
186+
stack.pop();
187+
else if(c == ']' && !stack.isEmpty() && stack.peek() == '[')
188+
stack.pop();
189+
else
190+
stack.push(c);// 没有匹配的就放进去
191+
}
192+
193+
return stack.isEmpty();
194+
}
195+
}
196+
```
197+
175198
### Python:
176199

177200
```python
@@ -555,3 +578,4 @@ impl Solution {
555578
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
556579
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
557580
</a>
581+

‎problems/0059.螺旋矩阵II.md‎

Lines changed: 57 additions & 18 deletions
Original file line numberDiff line numberDiff line change
@@ -715,26 +715,65 @@ object Solution {
715715
### C#:
716716

717717
```csharp
718-
public class Solution {
719-
public int[][] GenerateMatrix(int n) {
720-
int[][] answer = new int[n][];
721-
for(int i = 0; i < n; i++)
722-
answer[i] = new int[n];
723-
int start = 0;
724-
int end = n - 1;
725-
int tmp = 1;
726-
while(tmp < n * n)
718+
public int[][] GenerateMatrix(int n)
719+
{
720+
// 参考Carl的代码随想录里面C++的思路
721+
// https://www.programmercarl.com/0059.%E8%9E%BA%E6%97%8B%E7%9F%A9%E9%98%B5II.html#%E6%80%9D%E8%B7%AF
722+
int startX = 0, startY = 0; // 定义每循环一个圈的起始位置
723+
int loop = n / 2; // 每个圈循环几次,例如n为奇数3,那么loop = 1 只是循环一圈,矩阵中间的值需要单独处理
724+
int count = 1; // 用来给矩阵每个空格赋值
725+
int mid = n / 2; // 矩阵中间的位置,例如:n为3, 中间的位置就是(1,1),n为5,中间位置为(2, 2)
726+
int offset = 1;// 需要控制每一条边遍历的长度,每次循环右边界收缩一位
727+
728+
// 构建result二维数组
729+
int[][] result = new int[n][];
730+
for (int k = 0; k < n; k++)
731+
{
732+
result[k] = new int[n];
733+
}
734+
735+
int i = 0, j = 0; // [i,j]
736+
while (loop > 0)
737+
{
738+
i = startX;
739+
j = startY;
740+
// 四个For循环模拟转一圈
741+
// 第一排,从左往右遍历,不取最右侧的值(左闭右开)
742+
for (; j < n - offset; j++)
743+
{
744+
result[i][j] = count++;
745+
}
746+
// 右侧的第一列,从上往下遍历,不取最下面的值(左闭右开)
747+
for (; i < n - offset; i++)
748+
{
749+
result[i][j] = count++;
750+
}
751+
752+
// 最下面的第一行,从右往左遍历,不取最左侧的值(左闭右开)
753+
for (; j > startY; j--)
754+
{
755+
result[i][j] = count++;
756+
}
757+
758+
// 左侧第一列,从下往上遍历,不取最左侧的值(左闭右开)
759+
for (; i > startX; i--)
727760
{
728-
for(int i = start; i < end; i++) answer[start][i] = tmp++;
729-
for(int i = start; i < end; i++) answer[i][end] = tmp++;
730-
for(int i = end; i > start; i--) answer[end][i] = tmp++;
731-
for(int i = end; i > start; i--) answer[i][start] = tmp++;
732-
start++;
733-
end--;
734-
}
735-
if(n % 2 == 1) answer[n / 2][n / 2] = tmp;
736-
return answer;
761+
result[i][j] = count++;
762+
}
763+
// 第二圈开始的时候,起始位置要各自加1, 例如:第一圈起始位置是(0, 0),第二圈起始位置是(1, 1)
764+
startX++;
765+
startY++;
766+
767+
// offset 控制每一圈里每一条边遍历的长度
768+
offset++;
769+
loop--;
770+
}
771+
if (n % 2 == 1)
772+
{
773+
// n 为奇数
774+
result[mid][mid] = count;
737775
}
776+
return result;
738777
}
739778
```
740779

‎problems/0102.二叉树的层序遍历.md‎

Lines changed: 70 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -1231,6 +1231,47 @@ impl Solution {
12311231
}
12321232
```
12331233

1234+
#### C#:
1235+
1236+
```C# 199.二叉树的右视图
1237+
public class Solution
1238+
{
1239+
public IList<int> RightSideView(TreeNode root)
1240+
{
1241+
var result = new List<int>();
1242+
Queue<TreeNode> queue = new();
1243+
1244+
if (root != null)
1245+
{
1246+
queue.Enqueue(root);
1247+
}
1248+
while (queue.Count > 0)
1249+
{
1250+
int count = queue.Count;
1251+
int lastValue = count - 1;
1252+
for (int i = 0; i < count; i++)
1253+
{
1254+
var currentNode = queue.Dequeue();
1255+
if (i == lastValue)
1256+
{
1257+
result.Add(currentNode.val);
1258+
}
1259+
1260+
// lastValue == i == count -1 : left 先于 right 进入Queue
1261+
if (currentNode.left != null) queue.Enqueue(currentNode.left);
1262+
if (currentNode.right != null) queue.Enqueue(currentNode.right);
1263+
1264+
//// lastValue == i == 0: right 先于 left 进入Queue
1265+
// if(currentNode.right !=null ) queue.Enqueue(currentNode.right);
1266+
// if(currentNode.left !=null ) queue.Enqueue(currentNode.left);
1267+
}
1268+
}
1269+
1270+
return result;
1271+
}
1272+
}
1273+
```
1274+
12341275
## 637.二叉树的层平均值
12351276

12361277
[力扣题目链接](https://leetcode.cn/problems/average-of-levels-in-binary-tree/)
@@ -1558,6 +1599,35 @@ impl Solution {
15581599
}
15591600
```
15601601

1602+
#### C#:
1603+
1604+
```C# 二叉树的层平均值
1605+
public class Solution {
1606+
public IList<double> AverageOfLevels(TreeNode root) {
1607+
var result= new List<double>();
1608+
Queue<TreeNode> queue = new();
1609+
if(root !=null) queue.Enqueue(root);
1610+
1611+
while (queue.Count > 0)
1612+
{
1613+
int count = queue.Count;
1614+
double value=0;
1615+
for (int i = 0; i < count; i++)
1616+
{
1617+
var curentNode=queue.Dequeue();
1618+
value += curentNode.val;
1619+
if (curentNode.left!=null) queue.Enqueue(curentNode.left);
1620+
if (curentNode.right!=null) queue.Enqueue(curentNode.right);
1621+
}
1622+
result.Add(value/count);
1623+
}
1624+
1625+
return result;
1626+
}
1627+
}
1628+
1629+
```
1630+
15611631
## 429.N叉树的层序遍历
15621632

15631633
[力扣题目链接](https://leetcode.cn/problems/n-ary-tree-level-order-traversal/)

‎problems/0150.逆波兰表达式求值.md‎

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Original file line numberDiff line numberDiff line change
@@ -489,6 +489,67 @@ impl Solution {
489489
}
490490
```
491491

492+
### C:
493+
494+
```c
495+
int str_to_int(char *str) {
496+
// string转integer
497+
int num = 0, tens = 1;
498+
for (int i = strlen(str) - 1; i >= 0; i--) {
499+
if (str[i] == '-') {
500+
num *= -1;
501+
break;
502+
}
503+
num += (str[i] - '0') * tens;
504+
tens *= 10;
505+
}
506+
return num;
507+
}
508+
509+
int evalRPN(char** tokens, int tokensSize) {
510+
511+
int *stack = (int *)malloc(tokensSize * sizeof(int));
512+
assert(stack);
513+
int stackTop = 0;
514+
515+
for (int i = 0; i < tokensSize; i++) {
516+
char symbol = (tokens[i])[0];
517+
if (symbol < '0' && (tokens[i])[1] == '0円') {
518+
519+
// pop两个数字
520+
int num1 = stack[--stackTop];
521+
int num2 = stack[--stackTop];
522+
523+
// 计算结果
524+
int result;
525+
if (symbol == '+') {
526+
result = num1 + num2;
527+
} else if (symbol == '-') {
528+
result = num2 - num1;
529+
} else if (symbol == '/') {
530+
result = num2 / num1;
531+
} else {
532+
result = num1 * num2;
533+
}
534+
535+
// push回stack
536+
stack[stackTop++] = result;
537+
538+
} else {
539+
540+
// push数字进stack
541+
int num = str_to_int(tokens[i]);
542+
stack[stackTop++] = num;
543+
544+
}
545+
}
546+
547+
int result = stack[0];
548+
free(stack);
549+
return result;
550+
}
551+
```
552+
492553
<p align="center">
493554
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
494555
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>

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