Leetcode 4.21

1.罗马数字转整数

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;
    }
};

2.整数转罗马数字

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;
    }
};

3.三数之和

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;
    }
};

4.四数之和

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;
    }
};

5.电话号码的字母组合

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();
        }
    }
};

6.有效的括号

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;
    }
};