1. 回文排列
   

class Solution 
{
public:bool canPermutePalindrome(string s) {// 记录字符出现的次数int count[256] = {0};for(size_t i = 0; i < s.size(); ++i)++count[s[i]];// 记录字符出现次数为奇数的个数int flag = 0;for(size_t i = 0; i < 256; ++i)if(count[i] % 2 == 1)++flag;if(flag > 1)return false;return true;}
};

2. URL化
   

class Solution 
{
public:string replaceSpaces(string S, int length) {int front = length - 1;int back = front;// 遍历找出空格个数for(int i = 0; i < length; ++i){if(S[i] == ' '){back += 2;}}// 实际URL化后所需的空间大小S.resize(back + 1);while(front != back){if(S[front] != ' '){S[back--] = S[front];}else{S[back--] = '0';S[back--] = '2';S[back--] = '%';}--front;}return S;}
};

3. 配对交换
   

class Solution {
public:int exchangeBits(int num) {int num1 = 0b01010101010101010101010101010101;int num2 = 0b10101010101010101010101010101010;// 获取奇数位num1 &= num;// 获取偶数位num2 &= num;return (num1 << 1) + (num2 >> 1);}
};

4. 递归乘法
   

class Solution
{
public:// A * B = A + A * (B - 1) = A + A + A * (B - 2) = ....int multiply(int A, int B){if(B == 0)return 0;if(B == 1)return A;return A + multiply(A, B - 1);}
};

5. 阶乘尾数
   

/*
* 尾数0是乘10来的，10是从2*5来的
* [1,n]中因子5相对2会少些，找出[1,n]中因子5的个数，就是尾数0的个数
*/
class Solution 
{
public:int trailingZeroes(int n) {int count = 0;while(n >= 5){n /= 5;count += n;}return count;}
};

6. 二进制链表转整数
   
   

class Solution
{
public:int getDecimalValue(ListNode* head) {int num = 0;while(head != NULL){num = (num << 1) + head->val;head = head->next;}return num;}
};

7. 从链表中删去总和值为零的连续节点
   

class Solution {
public:ListNode* removeZeroSumSublists(ListNode* head){ListNode* newhead = new ListNode;newhead->next = head;ListNode* front = newhead;while (front){int sum = 0;ListNode* back = front->next;while (back){sum += back->val;if (sum == 0){ListNode* del = front->next;back = back->next;while (del != back){ListNode* tmp = del;del = del->next;// delete tmp;}front->next = back;}else{back = back->next;}}front = front->next;}head = newhead->next;delete newhead;return head;}
};

8. 括号的最大嵌套深度
   

class Solution {
public:int maxDepth(string s){int count = 0; // 记录当前深度int depth = 0; // 记录最深度for (char c : s){if (c == '('){++count;if(count > depth){++depth;}}else if (c == ')'){--count;}}return depth;}
};

9. 整理字符串
   
   

class Solution {
public:string makeGood(string s) {size_t i = 0;while(s.size() > 0 && i < s.size() - 1){if(abs(s[i] - s[i+1]) == 32){s.erase(i, 2);if(i > 0){--i;}continue;}++i;}return s;}
};

10. 奇偶树
    
    

class Solution {
public:bool isEvenOddTree(TreeNode* root){queue<TreeNode*> q;q.push(root);if((q.front()->val) % 2 == 0){return false;}int level = 0;int num = 1;while(!q.empty()){// v用于存储一层节点数据用于比较判断vector<int> v;while(num--){if(q.front()->left){q.push(q.front()->left);v.push_back(q.back()->val);}if(q.front()->right){q.push(q.front()->right);v.push_back(q.back()->val);}q.pop();}++level;num = v.size();if(!v.empty()){if(level % 2) // level 是奇{int prev = v[0];if(prev % 2 == 1){return false;}for(size_t i = 1; i < v.size(); ++i){if((v[i] % 2 != 0) || prev <= v[i]){return false;}prev = v[i];}}else // level 是偶{int prev = v[0];if(prev % 2 == 0){return false;}for(size_t i = 1; i < v.size(); ++i){if((v[i] % 2 != 1) ||  prev >= v[i]){return false;}prev = v[i];}}          }}    return true;}
};

11. 将句子排序
    
    

class Solution {
public:string sortSentence(string s){vector<string> vs;vs.resize(9);// 倒序遍历，找数字for(int i = s.size() - 1; i >= 0; --i){if(s[i] >= '0' && s[i] <= '9'){int j = i - 1;while(j >= 0 && s[j] != ' '){--j;}// 将对应编号的数字对应的字符串，放入数组vs[s[i] - '0' - 1] += string(s.begin() + j + 1, s.begin() + i);i = j;}}// 拼接到vs[0]for(size_t i = 1; i < 9; ++i){if(vs[i].size() > 0){vs[0] += (' ' + vs[i]);}}return vs[0];}
};

12. 最长和谐子序列
    
    

class Solution {
public:int findLHS(vector<int>& nums){map<int, int> m;// 去重 + 排序for (int e : nums){m[e]++;}auto it = m.begin();auto prev = it;++it;int max = 0;while (it != m.end()){if ((it->first - prev->first == 1) && (prev->second + it->second > max)){max = prev->second + it->second;}prev = it;++it;}return max;}
};