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Commit 467e6c7

Browse files
Merge branch 'youngyangyang04:master' into jian
2 parents d5e0827 + e8a10ab commit 467e6c7

8 files changed

+246
-43
lines changed

‎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/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">
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<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/0239.滑动窗口最大值.md

Lines changed: 32 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -890,6 +890,38 @@ public:
890890
};
891891
```
892892
893+
### C
894+
895+
```c
896+
int* maxSlidingWindow(int* nums, int numsSize, int k, int* returnSize) {
897+
*returnSize = numsSize - k + 1;
898+
int *res = (int*)malloc((*returnSize) * sizeof(int));
899+
assert(res);
900+
int *deque = (int*)malloc(numsSize * sizeof(int));
901+
assert(deque);
902+
int front = 0, rear = 0, idx = 0;
903+
904+
for (int i = 0 ; i < numsSize ; i++) {
905+
while (front < rear && deque[front] <= i - k) {
906+
front++;
907+
}
908+
909+
while (front < rear && nums[deque[rear - 1]] <= nums[i]) {
910+
rear--;
911+
}
912+
913+
deque[rear++] = i;
914+
915+
if (i >= k - 1) {
916+
res[idx++] = nums[deque[front]];
917+
}
918+
}
919+
920+
return res;
921+
}
922+
923+
```
924+
893925
<p align="center">
894926
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
895927
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>

‎problems/0459.重复的子字符串.md

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Original file line numberDiff line numberDiff line change
@@ -879,6 +879,52 @@ public int[] GetNext(string s)
879879
}
880880
```
881881

882+
### C
883+
884+
```c
885+
// 前缀表不减一
886+
int *build_next(char* s, int len) {
887+
888+
int *next = (int *)malloc(len * sizeof(int));
889+
assert(next);
890+
891+
// 初始化前缀表
892+
next[0] = 0;
893+
894+
// 构建前缀表表
895+
int i = 1, j = 0;
896+
while (i < len) {
897+
if (s[i] == s[j]) {
898+
j++;
899+
next[i] = j;
900+
i++;
901+
} else if (j > 0) {
902+
j = next[j - 1];
903+
} else {
904+
next[i] = 0;
905+
i++;
906+
}
907+
}
908+
return next;
909+
}
910+
911+
bool repeatedSubstringPattern(char* s) {
912+
913+
int len = strlen(s);
914+
int *next = build_next(s, len);
915+
bool result = false;
916+
917+
// 检查最小重复片段能否被长度整除
918+
if (next[len - 1]) {
919+
result = len % (len - next[len - 1]) == 0;
920+
}
921+
922+
free(next);
923+
return result;
924+
}
925+
926+
```
927+
882928
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<p align="center">
884930
<a href="https://programmercarl.com/other/kstar.html" target="_blank">

‎problems/0494.目标和.md

Lines changed: 11 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -670,18 +670,26 @@ class Solution:
670670

671671
# 创建二维动态规划数组,行表示选取的元素数量,列表示累加和
672672
dp = [[0] * (target_sum + 1) for _ in range(len(nums) + 1)]
673+
dp = [[0] * (target_sum + 1) for _ in range(len(nums))]
673674

674675
# 初始化状态
675676
dp[0][0] = 1
677+
if nums[0] <= target_sum:
678+
dp[0][nums[0]] = 1
679+
numZero = 0
680+
for i in range(len(nums)):
681+
if nums[i] == 0:
682+
numZero += 1
683+
dp[i][0] = int(math.pow(2, numZero))
676684

677685
# 动态规划过程
678-
for i in range(1, len(nums)+1):
686+
for i in range(1, len(nums)):
679687
for j in range(target_sum + 1):
680688
dp[i][j] = dp[i - 1][j] # 不选取当前元素
681689
if j >= nums[i - 1]:
682-
dp[i][j] += dp[i - 1][j - nums[i-1]] # 选取当前元素
690+
dp[i][j] += dp[i - 1][j - nums[i]] # 选取当前元素
683691

684-
return dp[len(nums)][target_sum] # 返回达到目标和的方案数
692+
return dp[len(nums)-1][target_sum] # 返回达到目标和的方案数
685693

686694

687695
```

‎problems/0518.零钱兑换II.md

Lines changed: 1 addition & 35 deletions
Original file line numberDiff line numberDiff line change
@@ -546,7 +546,7 @@ object Solution {
546546
}
547547
}
548548
```
549-
## C
549+
### C
550550

551551
```c
552552
int change(int amount, int* coins, int coinsSize) {
@@ -593,37 +593,3 @@ public class Solution
593593
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
594594
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
595595
</a>
596-
597-
----------
598-
599-
600-
601-
回归本题,动规五步曲来分析如下:
602-
603-
1. 确定dp数组以及下标的含义
604-
605-
dp[j]:凑成总金额j的货币组合数为dp[j]
606-
607-
2. 确定递推公式
608-
609-
dp[j] 就是所有的dp[j - coins[i]](考虑coins[i]的情况)相加。
610-
611-
所以递推公式:dp[j] += dp[j - coins[i]];
612-
613-
**这个递推公式大家应该不陌生了,我在讲解01背包题目的时候在这篇[494. 目标和](https://programmercarl.com/0494.目标和.html)中就讲解了,求装满背包有几种方法,公式都是:dp[j] += dp[j - nums[i]];**
614-
615-
3. dp数组如何初始化
616-
617-
首先dp[0]一定要为1,dp[0] = 1是 递归公式的基础。如果dp[0] = 0 的话,后面所有推导出来的值都是0了。
618-
619-
那么 dp[0] = 1 有没有含义,其实既可以说 凑成总金额0的货币组合数为1,也可以说 凑成总金额0的货币组合数为0,好像都没有毛病。
620-
621-
但题目描述中,也没明确说 amount = 0 的情况,结果应该是多少。
622-
623-
这里我认为题目描述还是要说明一下,因为后台测试数据是默认,amount = 0 的情况,组合数为1的。
624-
625-
下标非0的dp[j]初始化为0,这样累计加dp[j - coins[i]]的时候才不会影响真正的dp[j]
626-
627-
dp[0]=1还说明了一种情况:如果正好选了coins[i]后,也就是j-coins[i] == 0的情况表示这个硬币刚好能选,此时dp[0]为1表示只选coins[i]存在这样的一种选法。
628-
629-
----------------

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