Merge pull request #1043 from KingArthur0205/remote

添加 0343.整数拆分.md, 0096.不同的二叉搜索树.md, 0005.最长回文子串.md, 0143.重排链表.md, 0052.N皇后II.md, 0129.求根到叶子节点数字之和.md, 0209.长度最小的子数组.md  C语言解法
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程序员Carl
2022-02-03 00:56:44 +08:00
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7 changed files with 351 additions and 0 deletions

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@ -462,7 +462,92 @@ 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;
}
```
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;
}
```
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div> <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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@ -143,5 +143,65 @@ var totalNQueens = function(n) {
return count; 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;
}
```
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div> <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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@ -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] = 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];
}
}
return dp[n];
}
```
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div> <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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@ -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;
}
```
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div> <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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@ -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;
}
```
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@ -330,5 +330,55 @@ def min_sub_array_len(target, nums)
end 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;
}
```
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div> <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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@ -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];
}
```
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