From 59c95ff24d60ec7809aa80c632b3728291892077 Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Thu, 20 Jan 2022 08:29:02 +0000 Subject: [PATCH 1/9] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200343.=E6=95=B4?= =?UTF-8?q?=E6=95=B0=E6=8B=86=E5=88=86.md=20C=E8=AF=AD=E8=A8=80=E8=A7=A3?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0343.整数拆分.md | 35 +++++++++++++++++++++++++++++++++++ 1 file changed, 35 insertions(+) diff --git a/problems/0343.整数拆分.md b/problems/0343.整数拆分.md index 5d11f670..9882c6ea 100644 --- a/problems/0343.整数拆分.md +++ b/problems/0343.整数拆分.md @@ -271,5 +271,40 @@ var integerBreak = function(n) { }; ``` +C: +```c +//初始化DP数组 +int *initDP(int num) { + int* dp = (int*)malloc(sizeof(int) * (num + 1)); + int i; + for(i = 0; i < num + 1; ++i) { + dp[i] = 0; + } + return dp; +} + +//取三数最大值 +int max(int num1, int num2, int num3) { + int tempMax = num1 > num2 ? num1 : num2; + return tempMax > num3 ? tempMax : num3; +} + +int integerBreak(int n){ + int *dp = initDP(n); + //初始化dp[2]为1 + dp[2] = 1; + + int i; + for(i = 3; i <= n; ++i) { + int j; + for(j = 1; j < i - 1; ++j) { + //取得上次循环:dp[i]，原数相乘，或j*dp[]i-j] 三数中的最大值 + dp[i] = max(dp[i], j * (i - j), j * dp[i - j]); + } + } + return dp[n]; +} +``` + -----------------------
From 2b33b45170c22eb4b900bc85ff04ad5ce0a5c2e2 Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Sat, 22 Jan 2022 10:39:07 +0000 Subject: [PATCH 2/9] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200096.=E4=B8=8D?= =?UTF-8?q?=E5=90=8C=E7=9A=84=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91?= =?UTF-8?q?.md=20C=E8=AF=AD=E8=A8=80=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0096.不同的二叉搜索树.md | 27 +++++++++++++++++++++++ 1 file changed, 27 insertions(+) diff --git a/problems/0096.不同的二叉搜索树.md b/problems/0096.不同的二叉搜索树.md index d4b8d024..d6f7aca7 100644 --- a/problems/0096.不同的二叉搜索树.md +++ b/problems/0096.不同的二叉搜索树.md @@ -227,7 +227,34 @@ const numTrees =(n) => { }; ``` +C: +```c +//开辟dp数组 +int *initDP(int n) { + int *dp = (int *)malloc(sizeof(int) * (n + 1)); + int i; + for(i = 0; i <= n; ++i) + dp[i] = 0; + return dp; +} +int numTrees(int n){ + //开辟dp数组 + int *dp = initDP(n); + //将dp[0]设为1 + dp[0] = 1; + + int i, j; + for(i = 1; i <= n; ++i) { + for(j = 1; j <= i; ++j) { + //递推公式：dp[i] = d[i] + 根为j时左子树种类个数 * 根为j时右子树种类个数 + dp[i] += dp[j - 1] * dp[i - j]; + } + } + + return dp[n]; +} +``` -----------------------
From 718e67acef29a7f21f7914d0b1e175b2b52a4063 Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Sat, 22 Jan 2022 10:41:09 +0000 Subject: [PATCH 3/9] =?UTF-8?q?=E4=BF=AE=E6=94=B9=200096.=E4=B8=8D?= =?UTF-8?q?=E5=90=8C=E7=9A=84=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91?= =?UTF-8?q?.md=20C=E8=AF=AD=E8=A8=80=E8=A7=A3=E6=B3=95=E6=B3=A8=E9=87=8A?= =?UTF-8?q?=E9=94=99=E5=88=AB=E5=AD=97?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0096.不同的二叉搜索树.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0096.不同的二叉搜索树.md b/problems/0096.不同的二叉搜索树.md index d6f7aca7..d334a29c 100644 --- a/problems/0096.不同的二叉搜索树.md +++ b/problems/0096.不同的二叉搜索树.md @@ -247,7 +247,7 @@ int numTrees(int n){ int i, j; for(i = 1; i <= n; ++i) { for(j = 1; j <= i; ++j) { - //递推公式：dp[i] = d[i] + 根为j时左子树种类个数 * 根为j时右子树种类个数 + //递推公式：dp[i] = dp[i] + 根为j时左子树种类个数 * 根为j时右子树种类个数 dp[i] += dp[j - 1] * dp[i - j]; } } From fee948b2af53b7bfc583fbbe5c43cd98c04e462f Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Mon, 24 Jan 2022 08:57:02 +0000 Subject: [PATCH 4/9] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200005.=E6=9C=80?= =?UTF-8?q?=E9=95=BF=E5=9B=9E=E6=96=87=E5=AD=90=E4=B8=B2.md=20C=E8=AF=AD?= =?UTF-8?q?=E8=A8=80=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0005.最长回文子串.md | 49 +++++++++++++++++++++++++++++ 1 file changed, 49 insertions(+) diff --git a/problems/0005.最长回文子串.md b/problems/0005.最长回文子串.md index f204a607..588afc22 100644 --- a/problems/0005.最长回文子串.md +++ b/problems/0005.最长回文子串.md @@ -462,7 +462,56 @@ var longestPalindrome = function(s) { }; ``` +## C +动态规划: +```c +//初始化dp数组，全部初始为false +bool **initDP(int strLen) { + bool **dp = (bool **)malloc(sizeof(bool *) * strLen); + int i, j; + for(i = 0; i < strLen; ++i) { + dp[i] = (bool *)malloc(sizeof(bool) * strLen); + for(j = 0; j < strLen; ++j) + dp[i][j] = false; + } + return dp; +} +char * longestPalindrome(char * s){ + //求出字符串长度 + int strLen = strlen(s); + //初始化dp数组，元素初始化为false + bool **dp = initDP(strLen); + int maxLength = 0, left = 0, right = 0; + + //从下到上，从左到右遍历 + int i, j; + for(i = strLen - 1; i >= 0; --i) { + for(j = i; j < strLen; ++j) { + //若当前i与j所指字符一样 + if(s[i] == s[j]) { + //若i、j指向相邻字符或同一字符，则为回文字符串 + if(j - i <= 1) + dp[i][j] = true; + //若i+1与j-1所指字符串为回文字符串，则i、j所指字符串为回文字符串 + else if(dp[i + 1][j - 1]) + dp[i][j] = true; + } + //若新的字符串的长度大于之前的最大长度，进行更新 + if(dp[i][j] && j - i + 1 > maxLength) { + maxLength = j - i + 1; + left = i; + right = j; + } + } + } + //复制回文字符串，并返回 + char *ret = (char*)malloc(sizeof(char) * (maxLength + 1)); + memcpy(ret, s + left, maxLength); + ret[maxLength] = 0; + return ret; +} +``` -----------------------
From 3a46ffdad4cfcc2bc3695db93f4760b9ae70b4d2 Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Tue, 25 Jan 2022 12:54:50 +0000 Subject: [PATCH 5/9] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200005.=E6=9C=80?= =?UTF-8?q?=E9=95=BF=E5=9B=9E=E6=96=87=E5=AD=90=E4=B8=B2.md=20C=E8=AF=AD?= =?UTF-8?q?=E8=A8=80=E5=8F=8C=E6=8C=87=E9=92=88=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0005.最长回文子串.md | 36 +++++++++++++++++++++++++++++ 1 file changed, 36 insertions(+) diff --git a/problems/0005.最长回文子串.md b/problems/0005.最长回文子串.md index 588afc22..99458825 100644 --- a/problems/0005.最长回文子串.md +++ b/problems/0005.最长回文子串.md @@ -513,5 +513,41 @@ char * longestPalindrome(char * s){ } ``` +双指针: +```c +int left, maxLength; +void extend(char *str, int i, int j, int size) { + while(i >= 0 && j < size && str[i] == str[j]) { + //若当前子字符串长度大于最长的字符串长度，进行更新 + if(j - i + 1 > maxLength) { + maxLength = j - i + 1; + left = i; + } + //左指针左移，右指针右移。扩大搜索范围 + ++j, --i; + } +} + +char * longestPalindrome(char * s){ + left = right = maxLength = 0; + int size = strlen(s); + + int i; + for(i = 0; i < size; ++i) { + //长度为单数的子字符串 + extend(s, i, i, size); + //长度为双数的子字符串 + extend(s, i, i + 1, size); + } + + //复制子字符串 + char *subStr = (char *)malloc(sizeof(char) * (maxLength + 1)); + memcpy(subStr, s + left, maxLength); + subStr[maxLength] = 0; + + return subStr; +} +``` + -----------------------
From 9d09b11d56722748e00d3647cdcd228793baaa66 Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Wed, 26 Jan 2022 09:26:03 +0000 Subject: [PATCH 6/9] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200143.=E9=87=8D?= =?UTF-8?q?=E6=8E=92=E9=93=BE=E8=A1=A8.md=20C=E8=AF=AD=E8=A8=80=E8=A7=A3?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0143.重排链表.md | 68 +++++++++++++++++++++++++++++++++++ 1 file changed, 68 insertions(+) diff --git a/problems/0143.重排链表.md b/problems/0143.重排链表.md index 4ea9cb97..00622623 100644 --- a/problems/0143.重排链表.md +++ b/problems/0143.重排链表.md @@ -439,7 +439,75 @@ var reorderList = function(head, s = [], tmp) { } ``` +### C +方法三：反转链表 +```c +//翻转链表 +struct ListNode *reverseList(struct ListNode *head) { + if(!head) + return NULL; + struct ListNode *preNode = NULL, *curNode = head; + while(curNode) { + //创建tempNode记录curNode->next（即将被更新） + struct ListNode* tempNode = curNode->next; + //将curNode->next指向preNode + curNode->next = preNode; + //更新preNode为curNode + preNode = curNode; + //curNode更新为原链表中下一个元素 + curNode = tempNode; + } + return preNode; +} +void reorderList(struct ListNode* head){ + //slow用来截取到链表的中间节点(第一个链表的最后节点)，每次循环跳一个节点。fast用来辅助，每次循环跳两个节点 + struct ListNode *fast = head, *slow = head; + while(fast && fast->next && fast->next->next) { + //fast每次跳两个节点 + fast = fast->next->next; + //slow每次跳一个节点 + slow = slow->next; + } + //将slow->next后的节点翻转 + struct ListNode *sndLst = reverseList(slow->next); + //将第一个链表与第二个链表断开 + slow->next = NULL; + //因为插入从curNode->next开始，curNode刚开始已经head。所以fstList要从head->next开始 + struct ListNode *fstLst = head->next; + struct ListNode *curNode = head; + + int count = 0; + //当第一个链表和第二个链表中都有节点时循环 + while(sndLst && fstLst) { + //count为奇数，插入fstLst中的节点 + if(count % 2) { + curNode->next = fstLst; + fstLst = fstLst->next; + } + //count为偶数，插入sndList的节点 + else { + curNode->next = sndLst; + sndLst = sndLst->next; + } + //设置下一个节点 + curNode = curNode->next; + //更新count + ++count; + } + + //若两个链表fstList和sndLst中还有节点，将其放入链表 + if(fstLst) { + curNode->next = fstLst; + } + if(sndLst) { + curNode->next = sndLst; + } + + //返回链表 + return head; +} +``` ----------------------- From bb02fb957bfef34ffc7270e70b4cf41b829ed3a6 Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Wed, 26 Jan 2022 23:13:07 +0000 Subject: [PATCH 7/9] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200052.N=E7=9A=87?= =?UTF-8?q?=E5=90=8EII.md=20C=E8=AF=AD=E8=A8=80=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0052.N皇后II.md | 60 ++++++++++++++++++++++++++++++++++++++ 1 file changed, 60 insertions(+) diff --git a/problems/0052.N皇后II.md b/problems/0052.N皇后II.md index 78531798..67e439ca 100644 --- a/problems/0052.N皇后II.md +++ b/problems/0052.N皇后II.md @@ -143,5 +143,65 @@ var totalNQueens = function(n) { return count; }; ``` + +C +```c +//path[i]为在i行，path[i]列上存在皇后 +int *path; +int pathTop; +int answer; +//检查当前level行index列放置皇后是否合法 +int isValid(int index, int level) { + int i; + //updater为若斜角存在皇后，其所应在的列 + //用来检查左上45度是否存在皇后 + int lCornerUpdater = index - level; + //用来检查右上135度是否存在皇后 + int rCornerUpdater = index + level; + for(i = 0; i < pathTop; ++i) { + //path[i] == index检查index列是否存在皇后 + //检查斜角皇后：只要path[i] == updater，就说明当前位置不可放置皇后。 + //path[i] == lCornerUpdater检查左上角45度是否有皇后 + //path[i] == rCornerUpdater检查右上角135度是否有皇后 + if(path[i] == index || path[i] == lCornerUpdater || path[i] == rCornerUpdater) + return 0; + //更新updater指向下一行对应的位置 + ++lCornerUpdater; + --rCornerUpdater; + } + return 1; +} + +//回溯算法：level为当前皇后行数 +void backTracking(int n, int level) { + //若path中元素个数已经为n，则证明有一种解法。answer+1 + if(pathTop == n) { + ++answer; + return; + } + + int i; + for(i = 0; i < n; ++i) { + //若当前level行，i列是合法的放置位置。就将i放入path中 + if(isValid(i, level)) { + path[pathTop++] = i; + backTracking(n, level + 1); + //回溯 + --pathTop; + } + } +} + +int totalNQueens(int n){ + answer = 0; + pathTop = 0; + path = (int *)malloc(sizeof(int) * n); + + backTracking(n, 0); + + return answer; +} +``` + -----------------------
From 023ec6900a759e030712edf5362b65569935df75 Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Thu, 27 Jan 2022 13:43:43 +0000 Subject: [PATCH 8/9] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200129.=E6=B1=82?= =?UTF-8?q?=E6=A0=B9=E5=88=B0=E5=8F=B6=E5=AD=90=E8=8A=82=E7=82=B9=E6=95=B0?= =?UTF-8?q?=E5=AD=97=E4=B9=8B=E5=92=8C.md=20C=E8=AF=AD=E8=A8=80=E8=A7=A3?= =?UTF-8?q?=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../0129.求根到叶子节点数字之和.md | 26 +++++++++++++++++++ 1 file changed, 26 insertions(+) diff --git a/problems/0129.求根到叶子节点数字之和.md b/problems/0129.求根到叶子节点数字之和.md index d07d5244..980779c2 100644 --- a/problems/0129.求根到叶子节点数字之和.md +++ b/problems/0129.求根到叶子节点数字之和.md @@ -289,7 +289,33 @@ var sumNumbers = function(root) { }; ``` +C: +```c +//sum记录总和 +int sum; +void traverse(struct TreeNode *node, int val) { + //更新val为根节点到当前节点的和 + val = val * 10 + node->val; + //若当前节点为叶子节点，记录val + if(!node->left && !node->right) { + sum+=val; + return; + } + //若有左/右节点，遍历左/右节点 + if(node->left) + traverse(node->left, val); + if(node->right) + traverse(node->right, val); +} +int sumNumbers(struct TreeNode* root){ + sum = 0; + + traverse(root, 0); + + return sum; +} +``` -----------------------
From 7fe4ea3fbc3d93665e2ed5663a1c41c6f21d5874 Mon Sep 17 00:00:00 2001 From: Guanzhong Pan Date: Thu, 27 Jan 2022 19:32:30 +0000 Subject: [PATCH 9/9] =?UTF-8?q?Add=200209.=E9=95=BF=E5=BA=A6=E6=9C=80?= =?UTF-8?q?=E5=B0=8F=E7=9A=84=E5=AD=90=E6=95=B0=E7=BB=84.md=20C=E8=AF=AD?= =?UTF-8?q?=E8=A8=80=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0209.长度最小的子数组.md | 50 +++++++++++++++++++++++ 1 file changed, 50 insertions(+) diff --git a/problems/0209.长度最小的子数组.md b/problems/0209.长度最小的子数组.md index 17422ca0..dc1d9f18 100644 --- a/problems/0209.长度最小的子数组.md +++ b/problems/0209.长度最小的子数组.md @@ -330,5 +330,55 @@ def min_sub_array_len(target, nums) end ``` +C: +暴力解法: +```c +int minSubArrayLen(int target, int* nums, int numsSize){ + //初始化最小长度为INT_MAX + int minLength = INT_MAX; + int sum; + + int left, right; + for(left = 0; left < numsSize; ++left) { + //每次遍历都清零sum，计算当前位置后和>=target的子数组的长度 + sum = 0; + //从left开始，sum中添加元素 + for(right = left; right < numsSize; ++right) { + sum += nums[right]; + //若加入当前元素后，和大于target，则更新minLength + if(sum >= target) { + int subLength = right - left + 1; + minLength = minLength < subLength ? minLength : subLength; + } + } + } + //若minLength不为INT_MAX，则返回minLnegth + return minLength == INT_MAX ? 0 : minLength; +} +``` + +滑动窗口: +```c +int minSubArrayLen(int target, int* nums, int numsSize){ + //初始化最小长度为INT_MAX + int minLength = INT_MAX; + int sum = 0; + + int left = 0, right = 0; + //右边界向右扩展 + for(; right < numsSize; ++right) { + sum += nums[right]; + //当sum的值大于等于target时，保存长度，并且收缩左边界 + while(sum >= target) { + int subLength = right - left + 1; + minLength = minLength < subLength ? minLength : subLength; + sum -= nums[left++]; + } + } + //若minLength不为INT_MAX，则返回minLnegth + return minLength == INT_MAX ? 0 : minLength; +} +``` + -----------------------