From 5cb33c319b8652f2694fccc9db88127ce5f24563 Mon Sep 17 00:00:00 2001 From: Arthur Date: Wed, 20 Oct 2021 16:10:04 +0100 Subject: [PATCH 1/4] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200106.=E4=BB=8E?= =?UTF-8?q?=E4=B8=AD=E5=BA=8F=E4=B8=8E=E5=90=8E=E7=BB=AD=E9=81=8D=E5=8E=86?= =?UTF-8?q?=E6=9E=84=E9=80=A0=E4=BA=8C=E5=8F=89=E6=A0=91.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 --- ...序与后序遍历序列构造二叉树.md | 75 +++++++++++++++++++ 1 file changed, 75 insertions(+) diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md index 39b55d8c..9e0b2430 100644 --- a/problems/0106.从中序与后序遍历序列构造二叉树.md +++ b/problems/0106.从中序与后序遍历序列构造二叉树.md @@ -818,6 +818,81 @@ var buildTree = function(preorder, inorder) { }; ``` +## C +106 从中序与后序遍历序列构造二叉树 +```c +int linearSearch(int* arr, int arrSize, int key) { + int i; + for(i = 0; i < arrSize; i++) { + if(arr[i] == key) + return i; + } + return -1; +} + +struct TreeNode* buildTree(int* inorder, int inorderSize, int* postorder, int postorderSize){ + //若中序遍历数组中没有元素，则返回NULL + if(!inorderSize) + return NULL; + //创建一个新的结点，将node的val设置为后序遍历的最后一个元素 + struct TreeNode* node = (struct TreeNode*)malloc(sizeof(struct TreeNode)); + node->val = postorder[postorderSize - 1]; + + //通过线性查找找到中间结点在中序数组中的位置 + int index = linearSearch(inorder, inorderSize, postorder[postorderSize - 1]); + + //左子树数组大小为index + //右子树的数组大小为数组大小减index减1（减的1为中间结点） + int rightSize = inorderSize - index - 1; + node->left = buildTree(inorder, index, postorder, index); + node->right = buildTree(inorder + index + 1, rightSize, postorder + index, rightSize); + return node; +} +``` + +105 从前序与中序遍历序列构造二叉树 +```c +struct TreeNode* buildTree(int* preorder, int preorderSize, int* inorder, int inorderSize){ + // 递归结束条件：传入的数组大小为0 + if(!preorderSize) + return NULL; + + // 1.找到前序遍历数组的第一个元素， 创建结点。左右孩子设置为NULL。 + int rootValue = preorder[0]; + struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode)); + root->val = rootValue; + root->left = NULL; + root->right = NULL; + + // 2.若前序遍历数组的大小为1，返回该结点 + if(preorderSize == 1) + return root; + + // 3.根据该结点切割中序遍历数组，将中序遍历数组分割成左右两个数组。算出他们的各自大小 + int index; + for(index = 0; index < inorderSize; index++) { + if(inorder[index] == rootValue) + break; + } + int leftNum = index; + int rightNum = inorderSize - index - 1; + + int* leftInorder = inorder; + int* rightInorder = inorder + leftNum + 1; + + // 4.根据中序遍历数组左右数组的各子大小切割前序遍历数组。也分为左右数组 + int* leftPreorder = preorder+1; + int* rightPreorder = preorder + 1 + leftNum; + + // 5.递归进入左右数组，将返回的结果作为根结点的左右孩子 + root->left = buildTree(leftPreorder, leftNum, leftInorder, leftNum); + root->right = buildTree(rightPreorder, rightNum, rightInorder, rightNum); + + // 6.返回根节点 + return root; +} +``` + ----------------------- * 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 8f1fe3c1b747716d98f22e4f14b2e25e94a108d1 Mon Sep 17 00:00:00 2001 From: Arthur Date: Thu, 21 Oct 2021 15:10:36 +0100 Subject: [PATCH 2/4] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200206.=E7=BF=BB?= =?UTF-8?q?=E8=BD=AC=E9=93=BE=E8=A1=A8.md=20C=E8=AF=AD=E8=A8=80=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/0206.翻转链表.md | 39 +++++++++++++++++++++++++++++++++++ 1 file changed, 39 insertions(+) diff --git a/problems/0206.翻转链表.md b/problems/0206.翻转链表.md index 45196386..08495259 100644 --- a/problems/0206.翻转链表.md +++ b/problems/0206.翻转链表.md @@ -371,6 +371,45 @@ func reverse(pre: ListNode?, cur: ListNode?) -> ListNode? { } ``` +C: +双指针法： +```c +struct ListNode* reverseList(struct ListNode* head){ + //保存cur的下一个结点 + struct ListNode* temp; + //pre指针指向前一个当前结点的前一个结点 + struct ListNode* pre = NULL; + //用head代替cur，也可以再定义一个cur结点指向head。 + while(head) { + //保存下一个结点的位置 + temp = head->next; + //翻转操作 + head->next = pre; + //更新结点 + pre = head; + head = temp; + } + return pre; +} +``` + +递归法： +```c +struct ListNode* reverse(struct ListNode* pre, struct ListNode* cur) { + if(!cur) + return pre; + struct ListNode* temp = cur->next; + cur->next = pre; + //将cur作为pre传入下一层 + //将temp作为cur传入下一层，改变其指针指向当前cur + return reverse(cur, temp); +} + +struct ListNode* reverseList(struct ListNode* head){ + return reverse(NULL, head); +} +``` + ----------------------- * 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 712ab9aa00a830eb0ef6b09ec8b20f2a9050b7da Mon Sep 17 00:00:00 2001 From: Arthur Date: Thu, 21 Oct 2021 15:31:22 +0100 Subject: [PATCH 3/4] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200226.=E7=BF=BB?= =?UTF-8?q?=E8=BD=AC=E4=BA=8C=E5=8F=89=E6=A0=91.md=20C=E8=AF=AD=E8=A8=80?= =?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/0226.翻转二叉树.md | 45 ++++++++++++++++++++++++++++++++ 1 file changed, 45 insertions(+) diff --git a/problems/0226.翻转二叉树.md b/problems/0226.翻转二叉树.md index b69cad55..d27d6d80 100644 --- a/problems/0226.翻转二叉树.md +++ b/problems/0226.翻转二叉树.md @@ -473,6 +473,51 @@ var invertTree = function(root) { }; ``` +C: +递归法 +```c +struct TreeNode* invertTree(struct TreeNode* root){ + if(!root) + return NULL; + //交换结点的左右孩子（中） + struct TreeNode* temp = root->right; + root->right = root->left; + root->left = temp; + 左 + invertTree(root->left); + //右 + invertTree(root->right); + return root; +} +``` +迭代法：深度优先遍历 +```c +struct TreeNode* invertTree(struct TreeNode* root){ + if(!root) + return NULL; + //存储结点的栈 + struct TreeNode** stack = (struct TreeNode**)malloc(sizeof(struct TreeNode*) * 100); + int stackTop = 0; + //将根节点入栈 + stack[stackTop++] = root; + //若栈中还有元素（进行循环） + while(stackTop) { + //取出栈顶元素 + struct TreeNode* temp = stack[--stackTop]; + //交换结点的左右孩子 + struct TreeNode* tempNode = temp->right; + temp->right = temp->left; + temp->left = tempNode; + //若当前结点有左右孩子，将其入栈 + if(temp->right) + stack[stackTop++] = temp->right; + if(temp->left) + stack[stackTop++] = temp->left; + } + return root; +} +``` + ----------------------- * 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 fb7a575c05fe6a70d34a82fbaff9d634472e1e23 Mon Sep 17 00:00:00 2001 From: Arthur Date: Fri, 22 Oct 2021 16:28:50 +0100 Subject: [PATCH 4/4] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200344.=E5=8F=8D?= =?UTF-8?q?=E8=BD=AC=E5=AD=97=E7=AC=A6=E4=B8=B2.md=20C=E8=AF=AD=E8=A8=80?= =?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/0344.反转字符串.md | 12 ++++++++++++ 1 file changed, 12 insertions(+) diff --git a/problems/0344.反转字符串.md b/problems/0344.反转字符串.md index a5f18d58..b08c3f21 100644 --- a/problems/0344.反转字符串.md +++ b/problems/0344.反转字符串.md @@ -218,7 +218,19 @@ func reverseString(_ s: inout [Character]) { } ``` +C: +```c +void reverseString(char* s, int sSize){ + int left = 0; + int right = sSize - 1; + while(left < right) { + char temp = s[left]; + s[left++] = s[right]; + s[right--] = temp; + } +} +``` -----------------------