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https://github.com/youngyangyang04/leetcode-master.git
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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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@ -462,7 +462,92 @@ var longestPalindrome = function(s) {
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};
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```
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## C
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动态规划:
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```c
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//初始化dp数组,全部初始为false
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bool **initDP(int strLen) {
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bool **dp = (bool **)malloc(sizeof(bool *) * strLen);
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int i, j;
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for(i = 0; i < strLen; ++i) {
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dp[i] = (bool *)malloc(sizeof(bool) * strLen);
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for(j = 0; j < strLen; ++j)
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dp[i][j] = false;
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}
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return dp;
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}
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char * longestPalindrome(char * s){
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//求出字符串长度
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int strLen = strlen(s);
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//初始化dp数组,元素初始化为false
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bool **dp = initDP(strLen);
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int maxLength = 0, left = 0, right = 0;
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//从下到上,从左到右遍历
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int i, j;
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for(i = strLen - 1; i >= 0; --i) {
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for(j = i; j < strLen; ++j) {
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//若当前i与j所指字符一样
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if(s[i] == s[j]) {
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//若i、j指向相邻字符或同一字符,则为回文字符串
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if(j - i <= 1)
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dp[i][j] = true;
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//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
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else if(dp[i + 1][j - 1])
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dp[i][j] = true;
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}
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//若新的字符串的长度大于之前的最大长度,进行更新
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if(dp[i][j] && j - i + 1 > maxLength) {
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maxLength = j - i + 1;
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left = i;
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right = j;
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}
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}
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}
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//复制回文字符串,并返回
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char *ret = (char*)malloc(sizeof(char) * (maxLength + 1));
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memcpy(ret, s + left, maxLength);
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ret[maxLength] = 0;
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return ret;
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}
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```
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双指针:
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```c
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int left, maxLength;
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void extend(char *str, int i, int j, int size) {
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while(i >= 0 && j < size && str[i] == str[j]) {
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//若当前子字符串长度大于最长的字符串长度,进行更新
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if(j - i + 1 > maxLength) {
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maxLength = j - i + 1;
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left = i;
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}
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//左指针左移,右指针右移。扩大搜索范围
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++j, --i;
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}
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}
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char * longestPalindrome(char * s){
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left = right = maxLength = 0;
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int size = strlen(s);
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int i;
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for(i = 0; i < size; ++i) {
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//长度为单数的子字符串
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extend(s, i, i, size);
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//长度为双数的子字符串
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extend(s, i, i + 1, size);
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}
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//复制子字符串
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char *subStr = (char *)malloc(sizeof(char) * (maxLength + 1));
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memcpy(subStr, s + left, maxLength);
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subStr[maxLength] = 0;
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return subStr;
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}
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```
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-----------------------
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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>
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@ -143,5 +143,65 @@ var totalNQueens = function(n) {
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return count;
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};
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```
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C
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```c
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//path[i]为在i行,path[i]列上存在皇后
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int *path;
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int pathTop;
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int answer;
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//检查当前level行index列放置皇后是否合法
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int isValid(int index, int level) {
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int i;
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//updater为若斜角存在皇后,其所应在的列
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//用来检查左上45度是否存在皇后
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int lCornerUpdater = index - level;
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//用来检查右上135度是否存在皇后
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int rCornerUpdater = index + level;
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for(i = 0; i < pathTop; ++i) {
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//path[i] == index检查index列是否存在皇后
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//检查斜角皇后:只要path[i] == updater,就说明当前位置不可放置皇后。
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//path[i] == lCornerUpdater检查左上角45度是否有皇后
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//path[i] == rCornerUpdater检查右上角135度是否有皇后
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if(path[i] == index || path[i] == lCornerUpdater || path[i] == rCornerUpdater)
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return 0;
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//更新updater指向下一行对应的位置
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++lCornerUpdater;
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--rCornerUpdater;
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}
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return 1;
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}
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//回溯算法:level为当前皇后行数
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void backTracking(int n, int level) {
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//若path中元素个数已经为n,则证明有一种解法。answer+1
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if(pathTop == n) {
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++answer;
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return;
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}
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int i;
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for(i = 0; i < n; ++i) {
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//若当前level行,i列是合法的放置位置。就将i放入path中
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if(isValid(i, level)) {
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path[pathTop++] = i;
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backTracking(n, level + 1);
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//回溯
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--pathTop;
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}
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}
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}
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int totalNQueens(int n){
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answer = 0;
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pathTop = 0;
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path = (int *)malloc(sizeof(int) * n);
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backTracking(n, 0);
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return answer;
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}
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```
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-----------------------
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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>
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@ -227,7 +227,34 @@ const numTrees =(n) => {
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};
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```
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C:
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```c
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//开辟dp数组
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int *initDP(int n) {
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int *dp = (int *)malloc(sizeof(int) * (n + 1));
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int i;
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for(i = 0; i <= n; ++i)
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dp[i] = 0;
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return dp;
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}
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int numTrees(int n){
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//开辟dp数组
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int *dp = initDP(n);
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//将dp[0]设为1
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dp[0] = 1;
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int i, j;
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for(i = 1; i <= n; ++i) {
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for(j = 1; j <= i; ++j) {
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//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
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dp[i] += dp[j - 1] * dp[i - j];
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}
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}
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return dp[n];
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}
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```
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-----------------------
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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>
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@ -289,7 +289,33 @@ var sumNumbers = function(root) {
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};
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```
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C:
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```c
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//sum记录总和
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int sum;
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void traverse(struct TreeNode *node, int val) {
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//更新val为根节点到当前节点的和
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val = val * 10 + node->val;
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//若当前节点为叶子节点,记录val
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if(!node->left && !node->right) {
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sum+=val;
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return;
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}
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//若有左/右节点,遍历左/右节点
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if(node->left)
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traverse(node->left, val);
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if(node->right)
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traverse(node->right, val);
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}
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int sumNumbers(struct TreeNode* root){
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sum = 0;
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traverse(root, 0);
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return sum;
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}
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```
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-----------------------
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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>
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@ -439,7 +439,75 @@ var reorderList = function(head, s = [], tmp) {
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}
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```
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### C
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方法三:反转链表
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```c
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//翻转链表
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struct ListNode *reverseList(struct ListNode *head) {
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if(!head)
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return NULL;
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struct ListNode *preNode = NULL, *curNode = head;
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while(curNode) {
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//创建tempNode记录curNode->next(即将被更新)
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struct ListNode* tempNode = curNode->next;
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//将curNode->next指向preNode
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curNode->next = preNode;
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//更新preNode为curNode
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preNode = curNode;
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//curNode更新为原链表中下一个元素
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curNode = tempNode;
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}
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return preNode;
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}
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void reorderList(struct ListNode* head){
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//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
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struct ListNode *fast = head, *slow = head;
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while(fast && fast->next && fast->next->next) {
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//fast每次跳两个节点
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fast = fast->next->next;
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//slow每次跳一个节点
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slow = slow->next;
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}
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//将slow->next后的节点翻转
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struct ListNode *sndLst = reverseList(slow->next);
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//将第一个链表与第二个链表断开
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slow->next = NULL;
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//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
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struct ListNode *fstLst = head->next;
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struct ListNode *curNode = head;
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int count = 0;
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//当第一个链表和第二个链表中都有节点时循环
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while(sndLst && fstLst) {
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//count为奇数,插入fstLst中的节点
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if(count % 2) {
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curNode->next = fstLst;
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fstLst = fstLst->next;
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}
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//count为偶数,插入sndList的节点
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else {
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curNode->next = sndLst;
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sndLst = sndLst->next;
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}
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//设置下一个节点
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curNode = curNode->next;
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//更新count
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++count;
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}
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//若两个链表fstList和sndLst中还有节点,将其放入链表
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if(fstLst) {
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curNode->next = fstLst;
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}
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if(sndLst) {
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curNode->next = sndLst;
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}
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//返回链表
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return head;
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}
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```
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-----------------------
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@ -330,5 +330,55 @@ def min_sub_array_len(target, nums)
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end
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```
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C:
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暴力解法:
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```c
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int minSubArrayLen(int target, int* nums, int numsSize){
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//初始化最小长度为INT_MAX
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int minLength = INT_MAX;
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int sum;
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int left, right;
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for(left = 0; left < numsSize; ++left) {
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//每次遍历都清零sum,计算当前位置后和>=target的子数组的长度
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sum = 0;
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//从left开始,sum中添加元素
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for(right = left; right < numsSize; ++right) {
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sum += nums[right];
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//若加入当前元素后,和大于target,则更新minLength
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if(sum >= target) {
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int subLength = right - left + 1;
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minLength = minLength < subLength ? minLength : subLength;
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}
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}
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}
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//若minLength不为INT_MAX,则返回minLnegth
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return minLength == INT_MAX ? 0 : minLength;
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}
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```
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滑动窗口:
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```c
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int minSubArrayLen(int target, int* nums, int numsSize){
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//初始化最小长度为INT_MAX
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int minLength = INT_MAX;
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int sum = 0;
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int left = 0, right = 0;
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//右边界向右扩展
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for(; right < numsSize; ++right) {
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sum += nums[right];
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//当sum的值大于等于target时,保存长度,并且收缩左边界
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while(sum >= target) {
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int subLength = right - left + 1;
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minLength = minLength < subLength ? minLength : subLength;
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sum -= nums[left++];
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}
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}
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//若minLength不为INT_MAX,则返回minLnegth
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return minLength == INT_MAX ? 0 : minLength;
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}
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```
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-----------------------
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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>
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@ -271,5 +271,40 @@ var integerBreak = function(n) {
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};
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```
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C:
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```c
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//初始化DP数组
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int *initDP(int num) {
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int* dp = (int*)malloc(sizeof(int) * (num + 1));
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int i;
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for(i = 0; i < num + 1; ++i) {
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dp[i] = 0;
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}
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return dp;
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}
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//取三数最大值
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int max(int num1, int num2, int num3) {
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int tempMax = num1 > num2 ? num1 : num2;
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return tempMax > num3 ? tempMax : num3;
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}
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int integerBreak(int n){
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int *dp = initDP(n);
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//初始化dp[2]为1
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dp[2] = 1;
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int i;
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for(i = 3; i <= n; ++i) {
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int j;
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for(j = 1; j < i - 1; ++j) {
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//取得上次循环:dp[i],原数相乘,或j*dp[]i-j] 三数中的最大值
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dp[i] = max(dp[i], j * (i - j), j * dp[i - j]);
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}
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}
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return dp[n];
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}
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```
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-----------------------
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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>
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Reference in New Issue
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