用unordered_map去存罗马数字对应的数值，分情况讨论，把所有情况都列出来即可

class Solution {
public:unordered_map<char, int> mp = {{'I', 1},{'V', 5},{'X', 10},{'L', 50},{'C', 100},{'D', 500},{'M', 1000}};int romanToInt(string s) {int n = s.length();int sum = 0;        for (int i = 0; i < n; i++) {if ((i + 1 < n) && ((s[i] == 'I' && (s[i + 1] == 'V' || s[i + 1] == 'X')) ||(s[i] == 'X' && (s[i + 1] == 'L' || s[i + 1] == 'C')) ||(s[i] == 'C' && (s[i + 1] == 'D' || s[i + 1] == 'M')))) {sum = sum + mp[s[i + 1]] - mp[s[i]];i++;} else {sum += mp[s[i]];}}return sum;}
};

用vector<pair<int, string>>去存罗马数字symbol对应的数值val，如果val比num小，则num -= val; string ans + symbol
注意：
1.存储方式是vector<pair<int, string>> valueSymbols
2.遍历方式for (const auto &[value, symbol]: valueSymbols)

class Solution {
public:vector<pair<int, string>> valueSymbols = {{1000, "M"},{900,  "CM"},{500,  "D"},{400,  "CD"},{100,  "C"},{90,   "XC"},{50,   "L"},{40,   "XL"},{10,   "X"},{9,    "IX"},{5,    "V"},{4,    "IV"},{1,    "I"},};string intToRoman(int num) {string ans = "";for (const auto &[value, symbol]: valueSymbols) {while (num >= value) {num -= value;ans += symbol;}if (num == 0) {break;}}return ans;}
};

遍历数组，target - nums[i], 按两数之和的方法求解。注意：答案中不可以包含重复的三元组。
需要注意的点：去重！！！！

左右指针相同去重
while (l < r && nums[r] == nums[r - 1]) r–;
起始位置相同去重
if (i > 0 && nums[i] == nums[i - 1]) continue;

class Solution {
public:vector<vector<int>> threeSum(vector<int>& nums) {sort(nums.begin(), nums.end());vector<vector<int>> ans;int l = 0, r = nums.size() - 1, sum = 0;for (int i = 0; i < nums.size(); i++) {if (i > 0 && nums[i] == nums[i - 1]) continue;int l = i + 1, r = nums.size() - 1;int target = - nums[i];while (l < r) {if (nums[l] + nums[r] < target) {l++;} else if (nums[l] + nums[r] > target) {r--;} else {ans.push_back({nums[i], nums[l], nums[r]});while (l < r && nums[l] == nums[l + 1]) l++;while (l < r && nums[r] == nums[r - 1]) r--;r--;l++;}}}return ans;}
};

与题目3相同，注意题目：不重复的四元组！这个题可以遍历数组，固定一个元素，target - nums[i]，就变成了3数之和的解法。

class Solution {
public:vector<vector<int>> fourSum(vector<int>& nums, int target) {sort(nums.begin(), nums.end());vector<vector<int>> ans;int target_tmp;for (int i = 0; i < nums.size() - 3; i++) {if (i > 0 && nums[i] == nums[i - 1]) continue;for (int j = i + 1; j < nums.size() - 2; j++) {if (j > i + 1 && nums[j] == nums[j - 1]) continue;int l = j + 1, r = nums.size() - 1;target_tmp = target - nums[i] - nums[j];while (l < r) {if ((long)nums[l] + nums[r] < target_tmp) {l++;} else if ((long)nums[l] + nums[r] > target_tmp) {r--;} else {ans.push_back({nums[i], nums[j], nums[l], nums[r]});while (l < r && nums[l] == nums[l + 1]) l++;while (l < r && nums[r] == nums[r - 1]) r--;l++, r--;}}}}return ans;}
};

可以排列组合的题目都可以首先联想到全排列，递归时先判断是否符合递归终止条件，不符合则依次加当前电话号码位置对应的字母，加完后递归，

class Solution {
public:unordered_map<char, string> mp ={{'2', "abc"},{'3', "def"},{'4', "ghi"},{'5', "jkl"},{'6', "mno"},{'7', "pqrs"},{'8', "tuv"},{'9', "wxyz"}};vector<string> ans;string tmp;vector<string> letterCombinations(string digits) {if (digits.size() == 0) return ans;dfs(digits, mp, 0);return ans;}void dfs(string digits, unordered_map<char, string>& mp, int index) {if (tmp.size() == digits.size()) {ans.push_back(tmp);return;}for (auto ch : mp[digits[index]]) {tmp += ch;dfs(digits, mp, index + 1);tmp.pop_back();}}
};

碰到左括号入栈，右括号出栈看是否匹配

class Solution {
public:bool isValid(string s) {stack<char> valid;for (auto ch: s){if (ch == '(' || ch == '[' || ch == '{') {valid.push(ch);} else {if(valid.empty()) return false;auto left = valid.top();valid.pop();if ((ch == ')' && left != '(') || (ch == ']' && left != '[') || (ch == '}' && left != '{')) {return false;}}}return valid.empty() ? true : false;}
};