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Commit 588e42c

Browse files
Merge branch 'jian' of github.com:JianWangUCD/leetcode-master into jian
2 parents a66142c + a5973a5 commit 588e42c

13 files changed

+390
-66
lines changed

‎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/0070.爬楼梯.md

Lines changed: 5 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -130,8 +130,8 @@ public:
130130
};
131131
```
132132
133-
* 时间复杂度:$O(n)$
134-
* 空间复杂度:$O(n)$
133+
* 时间复杂度:O(n)
134+
* 空间复杂度:O(n)
135135
136136
当然依然也可以,优化一下空间复杂度,代码如下:
137137
@@ -154,8 +154,8 @@ public:
154154
};
155155
```
156156

157-
* 时间复杂度:$O(n)$
158-
* 空间复杂度:$O(1)$
157+
* 时间复杂度:O(n)
158+
* 空间复杂度:O(1)
159159

160160
后面将讲解的很多动规的题目其实都是当前状态依赖前两个,或者前三个状态,都可以做空间上的优化,**但我个人认为面试中能写出版本一就够了哈,清晰明了,如果面试官要求进一步优化空间的话,我们再去优化**。
161161

@@ -524,3 +524,4 @@ impl Solution {
524524
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
525525
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
526526
</a>
527+

‎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/0111.二叉树的最小深度.md

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -40,7 +40,7 @@
4040
本题依然是前序遍历和后序遍历都可以,前序求的是深度,后序求的是高度。
4141

4242
* 二叉树节点的深度:指从根节点到该节点的最长简单路径边的条数或者节点数(取决于深度从0开始还是从1开始)
43-
* 二叉树节点的高度:指从该节点到叶子节点的最长简单路径边的条数后者节点数(取决于高度从0开始还是从1开始)
43+
* 二叉树节点的高度:指从该节点到叶子节点的最长简单路径边的条数或者节点数(取决于高度从0开始还是从1开始)
4444

4545
那么使用后序遍历,其实求的是根节点到叶子节点的最小距离,就是求高度的过程,不过这个最小距离 也同样是最小深度。
4646

‎problems/0150.逆波兰表达式求值.md

Lines changed: 61 additions & 0 deletions
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"/>

‎problems/0225.用队列实现栈.md

Lines changed: 89 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -1277,6 +1277,95 @@ impl MyStack {
12771277
}
12781278
```
12791279

1280+
### C:
1281+
1282+
> C:单队列
1283+
1284+
```c
1285+
typedef struct Node {
1286+
int val;
1287+
struct Node *next;
1288+
} Node_t;
1289+
1290+
// 用单向链表实现queue
1291+
typedef struct {
1292+
Node_t *head;
1293+
Node_t *foot;
1294+
int size;
1295+
} MyStack;
1296+
1297+
MyStack* myStackCreate() {
1298+
MyStack *obj = (MyStack *)malloc(sizeof(MyStack));
1299+
assert(obj);
1300+
obj->head = NULL;
1301+
obj->foot = NULL;
1302+
obj->size = 0;
1303+
return obj;
1304+
}
1305+
1306+
void myStackPush(MyStack* obj, int x) {
1307+
1308+
Node_t *temp = (Node_t *)malloc(sizeof(Node_t));
1309+
assert(temp);
1310+
temp->val = x;
1311+
temp->next = NULL;
1312+
1313+
// 添加至queue末尾
1314+
if (obj->foot) {
1315+
obj->foot->next = temp;
1316+
} else {
1317+
obj->head = temp;
1318+
}
1319+
obj->foot = temp;
1320+
obj->size++;
1321+
}
1322+
1323+
int myStackPop(MyStack* obj) {
1324+
1325+
// 获取末尾元素
1326+
int target = obj->foot->val;
1327+
1328+
if (obj->head == obj->foot) {
1329+
free(obj->foot);
1330+
obj->head = NULL;
1331+
obj->foot = NULL;
1332+
} else {
1333+
1334+
Node_t *prev = obj->head;
1335+
// 移动至queue尾部节点前一个节点
1336+
while (prev->next != obj->foot) {
1337+
prev = prev->next;
1338+
}
1339+
1340+
free(obj->foot);
1341+
obj->foot = prev;
1342+
obj->foot->next = NULL;
1343+
}
1344+
1345+
obj->size--;
1346+
return target;
1347+
}
1348+
1349+
int myStackTop(MyStack* obj) {
1350+
return obj->foot->val;
1351+
}
1352+
1353+
bool myStackEmpty(MyStack* obj) {
1354+
return obj->size == 0;
1355+
}
1356+
1357+
void myStackFree(MyStack* obj) {
1358+
Node_t *curr = obj->head;
1359+
while (curr != NULL) {
1360+
Node_t *temp = curr->next;
1361+
free(curr);
1362+
curr = temp;
1363+
}
1364+
free(obj);
1365+
}
1366+
1367+
```
1368+
12801369
12811370
<p align="center">
12821371
<a href="https://programmercarl.com/other/kstar.html" target="_blank">

‎problems/0235.二叉搜索树的最近公共祖先.md

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -99,7 +99,7 @@ if (cur == NULL) return cur;
9999

100100
* 确定单层递归的逻辑
101101

102-
在遍历二叉搜索树的时候就是寻找区间[p->val, q->val](注意这里是左闭又闭)
102+
在遍历二叉搜索树的时候就是寻找区间[p->val, q->val](注意这里是左闭右闭)
103103

104104
那么如果 cur->val 大于 p->val,同时 cur->val 大于q->val,那么就应该向左遍历(说明目标区间在左子树上)。
105105

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