From 72df413ffcf9dbecf7a0101e82360317645a4881 Mon Sep 17 00:00:00 2001 From: ethaiyi9 Date: Tue, 26 Nov 2024 00:20:14 +0800 Subject: [PATCH 1/8] =?UTF-8?q?=E4=BF=AE=E6=94=B90111.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=9C=80=E5=B0=8F=E6=B7=B1=E5=BA=A6=20?= =?UTF-8?q?=E9=94=99=E5=AD=97?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0111.二叉树的最小深度.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0111.二叉树的最小深度.md b/problems/0111.二叉树的最小深度.md index cd7096ac..708e0532 100644 --- a/problems/0111.二叉树的最小深度.md +++ b/problems/0111.二叉树的最小深度.md @@ -40,7 +40,7 @@ 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。 * 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） -* 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） +* 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） 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。 From 455520bde6ba29d01eb9b43d3aba88b8f7d2dead Mon Sep 17 00:00:00 2001 From: Po1vre Date: Thu, 28 Nov 2024 15:54:24 +0800 Subject: [PATCH 2/8] =?UTF-8?q?fix:=2070=20=E7=88=AC=E6=A5=BC=E6=A2=AF?= =?UTF-8?q?=E5=88=A0=E5=8E=BB=E5=A4=8D=E6=9D=82=E5=BA=A6=E7=9A=84=E5=86=85?= =?UTF-8?q?=E8=81=94=E5=85=AC=E5=BC=8F=E7=AC=A6?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 仅有单对的‘$’，与其他文档同步，故删去 --- problems/0070.爬楼梯.md | 9 +++++---- 1 file changed, 5 insertions(+), 4 deletions(-) diff --git a/problems/0070.爬楼梯.md b/problems/0070.爬楼梯.md index a2f664a4..6a13a21c 100644 --- a/problems/0070.爬楼梯.md +++ b/problems/0070.爬楼梯.md @@ -130,8 +130,8 @@ public: }; ``` -* 时间复杂度：$O(n)$ -* 空间复杂度：$O(n)$ +* 时间复杂度：O(n) +* 空间复杂度：O(n) 当然依然也可以，优化一下空间复杂度，代码如下： @@ -154,8 +154,8 @@ public: }; ``` -* 时间复杂度：$O(n)$ -* 空间复杂度：$O(1)$ +* 时间复杂度：O(n) +* 空间复杂度：O(1) 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**。 @@ -524,3 +524,4 @@ impl Solution { + From 6776ecc172511158d6f516721afd8bfb91429014 Mon Sep 17 00:00:00 2001 From: Po1vre Date: Fri, 6 Dec 2024 14:37:45 +0800 Subject: [PATCH 3/8] =?UTF-8?q?docs:=200518=E9=9B=B6=E9=92=B1=E5=85=91?= =?UTF-8?q?=E6=8D=A2=E2=85=A1=20=E4=BF=AE=E6=94=B9=E6=A0=87=E7=AD=BE?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0518.零钱兑换II.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/problems/0518.零钱兑换II.md b/problems/0518.零钱兑换II.md index 0d35fb7c..1921866e 100644 --- a/problems/0518.零钱兑换II.md +++ b/problems/0518.零钱兑换II.md @@ -397,7 +397,7 @@ object Solution { } } ``` -## C +### C ```c int change(int amount, int* coins, int coinsSize) { @@ -444,3 +444,4 @@ public class Solution + From fb9186fc79c233a9cf06f9e5ecc3070f1d1c3332 Mon Sep 17 00:00:00 2001 From: C_W Date: Thu, 12 Dec 2024 12:15:00 +1100 Subject: [PATCH 4/8] =?UTF-8?q?=E6=B7=BB=E5=8A=A00459=E9=87=8D=E5=A4=8D?= =?UTF-8?q?=E7=9A=84=E5=AD=90=E5=AD=97=E7=AC=A6=E4=B8=B2=20C=20=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0459.重复的子字符串.md | 46 ++++++++++++++++++++++++++ 1 file changed, 46 insertions(+) diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md index 2be8922b..de0e6e4d 100644 --- a/problems/0459.重复的子字符串.md +++ b/problems/0459.重复的子字符串.md @@ -879,6 +879,52 @@ public int[] GetNext(string s) } ``` +### C + +```c +// 前缀表不减一 +int *build_next(char* s, int len) { + + int *next = (int *)malloc(len * sizeof(int)); + assert(next); + + // 初始化前缀表 + next[0] = 0; + + // 构建前缀表表 + int i = 1, j = 0; + while (i < len) { + if (s[i] == s[j]) { + j++; + next[i] = j; + i++; + } else if (j > 0) { + j = next[j - 1]; + } else { + next[i] = 0; + i++; + } + } + return next; +} + +bool repeatedSubstringPattern(char* s) { + + int len = strlen(s); + int *next = build_next(s, len); + bool result = false; + + // 检查最小重复片段能否被长度整除 + if (next[len - 1]) { + result = len % (len - next[len - 1]) == 0; + } + + free(next); + return result; +} + +``` +

From d6f7f3adbcd2532fafe6ffc06efc4e3d01f8d1ee Mon Sep 17 00:00:00 2001 From: C_W Date: Thu, 12 Dec 2024 22:32:25 +1100 Subject: [PATCH 5/8] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200225.=E7=94=A8?= =?UTF-8?q?=E9=98=9F=E5=88=97=E5=AE=9E=E7=8E=B0=E6=A0=88.md=20C=20?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0225.用队列实现栈.md | 89 +++++++++++++++++++++++++++++ 1 file changed, 89 insertions(+) diff --git a/problems/0225.用队列实现栈.md b/problems/0225.用队列实现栈.md index f0fe3a3c..73d9db1b 100644 --- a/problems/0225.用队列实现栈.md +++ b/problems/0225.用队列实现栈.md @@ -1277,6 +1277,95 @@ impl MyStack { } ``` +### C: + +> C：单队列 + +```c +typedef struct Node { + int val; + struct Node *next; +} Node_t; + +// 用单向链表实现queue +typedef struct { + Node_t *head; + Node_t *foot; + int size; +} MyStack; + +MyStack* myStackCreate() { + MyStack *obj = (MyStack *)malloc(sizeof(MyStack)); + assert(obj); + obj->head = NULL; + obj->foot = NULL; + obj->size = 0; + return obj; +} + +void myStackPush(MyStack* obj, int x) { + + Node_t *temp = (Node_t *)malloc(sizeof(Node_t)); + assert(temp); + temp->val = x; + temp->next = NULL; + + // 添加至queue末尾 + if (obj->foot) { + obj->foot->next = temp; + } else { + obj->head = temp; + } + obj->foot = temp; + obj->size++; +} + +int myStackPop(MyStack* obj) { + + // 获取末尾元素 + int target = obj->foot->val; + + if (obj->head == obj->foot) { + free(obj->foot); + obj->head = NULL; + obj->foot = NULL; + } else { + + Node_t *prev = obj->head; + // 移动至queue尾部节点前一个节点 + while (prev->next != obj->foot) { + prev = prev->next; + } + + free(obj->foot); + obj->foot = prev; + obj->foot->next = NULL; + } + + obj->size--; + return target; +} + +int myStackTop(MyStack* obj) { + return obj->foot->val; +} + +bool myStackEmpty(MyStack* obj) { + return obj->size == 0; +} + +void myStackFree(MyStack* obj) { + Node_t *curr = obj->head; + while (curr != NULL) { + Node_t *temp = curr->next; + free(curr); + curr = temp; + } + free(obj); +} + +``` +

From d66e733d6c66b37baec8acd0d42f8161a07bf8fc Mon Sep 17 00:00:00 2001 From: C_W Date: Thu, 12 Dec 2024 22:33:13 +1100 Subject: [PATCH 6/8] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200150.=E9=80=86?= =?UTF-8?q?=E6=B3=A2=E5=85=B0=E8=A1=A8=E8=BE=BE=E5=BC=8F=E6=B1=82=E5=80=BC?= =?UTF-8?q?.md=20C=20=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0150.逆波兰表达式求值.md | 61 +++++++++++++++++++++++ 1 file changed, 61 insertions(+) diff --git a/problems/0150.逆波兰表达式求值.md b/problems/0150.逆波兰表达式求值.md index 7d4031d7..0f1e9c23 100644 --- a/problems/0150.逆波兰表达式求值.md +++ b/problems/0150.逆波兰表达式求值.md @@ -502,6 +502,67 @@ impl Solution { } ``` +### C: + +```c +int str_to_int(char *str) { + // string转integer + int num = 0, tens = 1; + for (int i = strlen(str) - 1; i >= 0; i--) { + if (str[i] == '-') { + num *= -1; + break; + } + num += (str[i] - '0') * tens; + tens *= 10; + } + return num; +} + +int evalRPN(char** tokens, int tokensSize) { + + int *stack = (int *)malloc(tokensSize * sizeof(int)); + assert(stack); + int stackTop = 0; + + for (int i = 0; i < tokensSize; i++) { + char symbol = (tokens[i])[0]; + if (symbol < '0' && (tokens[i])[1] == '\0') { + + // pop两个数字 + int num1 = stack[--stackTop]; + int num2 = stack[--stackTop]; + + // 计算结果 + int result; + if (symbol == '+') { + result = num1 + num2; + } else if (symbol == '-') { + result = num2 - num1; + } else if (symbol == '/') { + result = num2 / num1; + } else { + result = num1 * num2; + } + + // push回stack + stack[stackTop++] = result; + + } else { + + // push数字进stack + int num = str_to_int(tokens[i]); + stack[stackTop++] = num; + + } + } + + int result = stack[0]; + free(stack); + return result; +} +``` +

From 739e3f891ce199d7258fa4a192b7fe3dd10a9251 Mon Sep 17 00:00:00 2001 From: C_W Date: Fri, 13 Dec 2024 11:43:42 +1100 Subject: [PATCH 7/8] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200239.=E6=BB=91?= =?UTF-8?q?=E5=8A=A8=E7=AA=97=E5=8F=A3=E6=9C=80=E5=A4=A7=E5=80=BC.md=20C?= =?UTF-8?q?=20=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0239.滑动窗口最大值.md | 32 ++++++++++++++++++++++++++ 1 file changed, 32 insertions(+) diff --git a/problems/0239.滑动窗口最大值.md b/problems/0239.滑动窗口最大值.md index caa24d8d..9bb3494d 100644 --- a/problems/0239.滑动窗口最大值.md +++ b/problems/0239.滑动窗口最大值.md @@ -890,6 +890,38 @@ public: }; ``` +### C + +```c +int* maxSlidingWindow(int* nums, int numsSize, int k, int* returnSize) { + *returnSize = numsSize - k + 1; + int *res = (int*)malloc((*returnSize) * sizeof(int)); + assert(res); + int *deque = (int*)malloc(numsSize * sizeof(int)); + assert(deque); + int front = 0, rear = 0, idx = 0; + + for (int i = 0 ; i < numsSize ; i++) { + while (front < rear && deque[front] <= i - k) { + front++; + } + + while (front < rear && nums[deque[rear - 1]] <= nums[i]) { + rear--; + } + + deque[rear++] = i; + + if (i >= k - 1) { + res[idx++] = nums[deque[front]]; + } + } + + return res; +} + +``` +

From 3ce802897f9cf19ef158cb197f031c1cf3cb8baf Mon Sep 17 00:00:00 2001 From: Murphy Tian Date: Mon, 16 Dec 2024 15:38:43 +0800 Subject: [PATCH 8/8] [Fix][DP][Target Sum] python 2D version align with the dp equation --- problems/0494.目标和.md | 14 +++++++++++--- 1 file changed, 11 insertions(+), 3 deletions(-) diff --git a/problems/0494.目标和.md b/problems/0494.目标和.md index dda3ad75..eef5ceb6 100644 --- a/problems/0494.目标和.md +++ b/problems/0494.目标和.md @@ -671,18 +671,26 @@ class Solution: # 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 dp = [[0] * (target_sum + 1) for _ in range(len(nums) + 1)] + dp = [[0] * (target_sum + 1) for _ in range(len(nums))] # 初始化状态 dp[0][0] = 1 + if nums[0] <= target_sum: + dp[0][nums[0]] = 1 + numZero = 0 + for i in range(len(nums)): + if nums[i] == 0: + numZero += 1 + dp[i][0] = int(math.pow(2, numZero)) # 动态规划过程 - for i in range(1, len(nums) + 1): + for i in range(1, len(nums)): for j in range(target_sum + 1): dp[i][j] = dp[i - 1][j] # 不选取当前元素 if j >= nums[i - 1]: - dp[i][j] += dp[i - 1][j - nums[i - 1]] # 选取当前元素 + dp[i][j] += dp[i - 1][j - nums[i]] # 选取当前元素 - return dp[len(nums)][target_sum] # 返回达到目标和的方案数 + return dp[len(nums)-1][target_sum] # 返回达到目标和的方案数 ```