From 20976616179c3960eb14dd3661b5dd8469cda481 Mon Sep 17 00:00:00 2001 From: Arthur Date: Sun, 31 Oct 2021 17:04:12 +0100 Subject: [PATCH 1/6] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=201002.=E6=9F=A5?= =?UTF-8?q?=E6=89=BE=E5=B8=B8=E7=94=A8=E5=AD=97=E7=AC=A6.md=20C=E8=AF=AD?= =?UTF-8?q?=E8=A8=80=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1002.查找常用字符.md | 52 +++++++++++++++++++++++++++++ 1 file changed, 52 insertions(+) diff --git a/problems/1002.查找常用字符.md b/problems/1002.查找常用字符.md index 44a02ceb..a8450f29 100644 --- a/problems/1002.查找常用字符.md +++ b/problems/1002.查找常用字符.md @@ -331,6 +331,58 @@ func commonChars(_ words: [String]) -> [String] { } ``` +C: +```c +void updateHashTable(int* hashTableOne, int* hashTableTwo) { + int i; + for(i = 0; i < 26; i++) { + hashTableOne[i] = hashTableOne[i] < hashTableTwo[i] ? hashTableOne[i] : hashTableTwo[i]; + } +} + +char ** commonChars(char ** words, int wordsSize, int* returnSize){ + //用来统计所有字母出现的最小频率 + int hashTable[26] = { 0 }; + //初始化返回的char**数组以及返回数组长度 + *returnSize = 0; + char** ret = (char**)malloc(sizeof(char*) * 100); + + //如果输入数组长度为0，则返回NULL + if(!wordsSize) + return NULL; + + int i; + //更新第一个单词的字母频率 + for(i = 0; i < strlen(words[0]); i++) + hashTable[words[0][i] - 'a']++; + //更新从第二个单词开始的字母频率 + for(i = 1; i < wordsSize; i++) { + //创建新的哈希表，记录新的单词的字母频率 + int newHashTable[26] = { 0 }; + int j; + for(j = 0; j < strlen(words[i]); j++) { + newHashTable[words[i][j] - 'a']++; + } + //更新原哈希表 + updateHashTable(hashTable, newHashTable); + } + + //将哈希表中的字符变为字符串放入ret中 + for(i = 0; i < 26; i++) { + if(hashTable[i]) { + int j; + for(j = 0; j < hashTable[i]; j++) { + char* tempString = (char*)malloc(sizeof(char) * 2); + tempString[0] = i + 'a'; + tempString[1] = '\0'; + ret[(*returnSize)++] = tempString; + } + } + } + return ret; +} +``` + ----------------------- * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw) * B站视频：[代码随想录](https://space.bilibili.com/525438321) From d47af5291680d2b2ef0604e5780089c4c9a817b2 Mon Sep 17 00:00:00 2001 From: Arthur Date: Sun, 31 Oct 2021 17:09:17 +0100 Subject: [PATCH 2/6] =?UTF-8?q?=E6=9B=B4=E6=96=B0=201002.=E6=9F=A5?= =?UTF-8?q?=E6=89=BE=E5=B8=B8=E7=94=A8=E5=AD=97=E7=AC=A6.md=20C=E8=AF=AD?= =?UTF-8?q?=E8=A8=80=E7=89=88=E6=9C=AC=E6=B3=A8=E9=87=8A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1002.查找常用字符.md | 1 + 1 file changed, 1 insertion(+) diff --git a/problems/1002.查找常用字符.md b/problems/1002.查找常用字符.md index a8450f29..5a8d1093 100644 --- a/problems/1002.查找常用字符.md +++ b/problems/1002.查找常用字符.md @@ -333,6 +333,7 @@ func commonChars(_ words: [String]) -> [String] { C: ```c +//若两个哈希表定义为char数组（每个单词的最大长度不会超过100，因此可以用char表示），可以提高时间和空间效率 void updateHashTable(int* hashTableOne, int* hashTableTwo) { int i; for(i = 0; i < 26; i++) { From 0d5c3c34de875f845e324ed1772ec3e0db521bbd Mon Sep 17 00:00:00 2001 From: ArthurP Date: Wed, 3 Nov 2021 23:20:18 +0100 Subject: [PATCH 3/6] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200110.=E5=B9=B3?= =?UTF-8?q?=E8=A1=A1=E4=BA=8C=E5=8F=89=E6=A0=91.md=20C=E8=AF=AD=E8=A8=80?= =?UTF-8?q?=E9=80=92=E5=BD=92=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0110.平衡二叉树.md | 30 ++++++++++++++++++++++++++++++ 1 file changed, 30 insertions(+) diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index 43a49758..e22a12b3 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -619,6 +619,36 @@ var isBalanced = function(root) { }; ``` +##C +递归法： +```c +int getDepth(struct TreeNode* node) { + //如果结点不存在，返回0 + if(!node) + return 0; + //求出右子树深度 + int rightDepth = getDepth(node->right); + //求出左子树深度 + int leftDepth = getDepth(node->left); + //返回左右子树中的较大值+1 + return rightDepth > leftDepth ? rightDepth + 1 : leftDepth + 1; +} + +bool isBalanced(struct TreeNode* root) { + //递归结束条件为：传入结点为NULL，返回True + if(!root) + return 1; + //求出左右子树的深度 + int leftDepth = getDepth(root->left); + int rightDepth = getDepth(root->right); + int diff; + //若左右子树绝对值差距大于1，返回False + if((diff = leftDepth - rightDepth) > 1 || diff < -1) + return 0; + //检查左右子树是否为平衡二叉树 + return isBalanced(root->right) && isBalanced(root->left); +} +``` ----------------------- * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw) From ad19a88ecf7559b2634f236c3a88a3ed15e37388 Mon Sep 17 00:00:00 2001 From: Arthur Date: Thu, 4 Nov 2021 10:21:03 +0000 Subject: [PATCH 4/6] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200110.=E5=B9=B3?= =?UTF-8?q?=E8=A1=A1=E4=BA=8C=E5=8F=89=E6=A0=91.md=20=E6=B3=A8=E9=87=8A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0110.平衡二叉树.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index 677fbc32..2e8c239f 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -635,7 +635,7 @@ var isBalanced = function(root) { }; ``` -##C +## C 递归法： ```c int getDepth(struct TreeNode* node) { From 18ba70884914ed10944da16d60bba1fda610262f Mon Sep 17 00:00:00 2001 From: Arthur Date: Thu, 4 Nov 2021 10:56:41 +0000 Subject: [PATCH 5/6] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200110.=E5=B9=B3?= =?UTF-8?q?=E8=A1=A1=E4=BA=8C=E5=8F=89=E6=A0=91.md=20C=E8=AF=AD=E8=A8=80?= =?UTF-8?q?=E8=BF=AD=E4=BB=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0110.平衡二叉树.md | 63 ++++++++++++++++++++++++++++++++ 1 file changed, 63 insertions(+) diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index 2e8c239f..f3d70a5c 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -666,6 +666,69 @@ bool isBalanced(struct TreeNode* root) { } ``` +迭代法： +```c +//计算结点深度 +int getDepth(struct TreeNode* node) { + struct TreeNode** stack = (struct TreeNode**)malloc(sizeof(struct TreeNode*) * 10000); + int stackTop = 0; + if(node) + stack[stackTop++] = node; + int result = 0; + int depth = 0; + + while(stackTop) { + struct TreeNode* tempNode = stack[--stackTop]; + if(tempNode) { + depth++; + stack[stackTop++] = tempNode; + stack[stackTop++] = NULL; + if(tempNode->left) + stack[stackTop++] = tempNode->left; + if(tempNode->right) + stack[stackTop++] = tempNode->right; + result = result > depth ? result : depth; + } + else { + tempNode = stack[--stackTop]; + depth--; + } + } + + return result; +} + +bool isBalanced(struct TreeNode* root){ + //开辟栈空间 + struct TreeNode** stack = (struct TreeNode**)malloc(sizeof(struct TreeNode*) * 10000); + int stackTop = 0; + + //若根节点不存在，返回True + if(!root) + return 1; + + //将根节点入栈 + stack[stackTop++] = root; + //当栈中有元素时，进行遍历 + while(stackTop) { + //将栈顶元素出栈 + struct TreeNode* node = stack[--stackTop]; + //计算左右子树的深度 + int diff = getDepth(node->right) - getDepth(node->left); + //若深度的绝对值大于1，返回False + if(diff > 1 || diff < -1) + return 0; + //如果栈顶结点有左右结点，将左右结点入栈 + if(node->left) + stack[stackTop++] = node->left; + if(node->right) + stack[stackTop++] = node->right; + } + //若二叉树遍历结束后没有返回False，则返回True + return 1; +} +``` + ----------------------- * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw) * B站视频：[代码随想录](https://space.bilibili.com/525438321) From a1004808796f6b97bd86753e1b9d859ebab4ff09 Mon Sep 17 00:00:00 2001 From: ArthurP Date: Sat, 6 Nov 2021 10:51:37 +0100 Subject: [PATCH 6/6] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200110.=E5=B9=B3?= =?UTF-8?q?=E8=A1=A1=E4=BA=8C=E5=8F=89=E6=A0=91.md=20C=E8=AF=AD=E8=A8=80?= =?UTF-8?q?=E8=BF=AD=E4=BB=A3=E6=B3=95=E6=B3=A8=E9=87=8A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0110.平衡二叉树.md | 9 +++++++++ 1 file changed, 9 insertions(+) diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index f3d70a5c..b0ecf44a 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -670,26 +670,35 @@ bool isBalanced(struct TreeNode* root) { ```c //计算结点深度 int getDepth(struct TreeNode* node) { + //开辟栈空间 struct TreeNode** stack = (struct TreeNode**)malloc(sizeof(struct TreeNode*) * 10000); int stackTop = 0; + //若传入结点存在，将其入栈。若不存在，函数直接返回0 if(node) stack[stackTop++] = node; int result = 0; int depth = 0; + //当栈中有元素时，进行迭代遍历 while(stackTop) { + //取出栈顶元素 struct TreeNode* tempNode = stack[--stackTop]; + //若栈顶元素非NULL，则将深度+1 if(tempNode) { depth++; + //将栈顶元素再次入栈，添加NULL表示此结点已被遍历 stack[stackTop++] = tempNode; stack[stackTop++] = NULL; + //若栈顶元素有左右孩子，则将孩子结点入栈 if(tempNode->left) stack[stackTop++] = tempNode->left; if(tempNode->right) stack[stackTop++] = tempNode->right; + //更新结果 result = result > depth ? result : depth; } else { + //若为NULL，则代表当前结点已被遍历，深度-1 tempNode = stack[--stackTop]; depth--; }