365 days challenge LeetCode1000 questions - Day 046 Generate a string with odd number of each character + add two numbers + valid parentheses

1374. 生成每种字符都是奇数个的字符串

代码实现（首刷自解）

class Solution {
    
public:
    string generateTheString(int n) {
    
        string ans = "";
        if (n % 2 == 0)  {
    
            ans = "z";
            for (int i = 0; i < n - 1; i++) {
    
                ans.append("a");
            }
        }
        else {
    
            for (int i = 0; i < n; i++) {
    
                ans.append("a");
            }
        }
        return ans;
    }
};

2. 两数相加

代码实现（首刷自解）

/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next(next) {} * }; */
class Solution {
    
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
    
        int flag = 0;
        ListNode* pl1 = l1, *pl2 = l2;
        ListNode* ans = new ListNode();
        ListNode* curNode = ans;
        int curVal = 0;
        while (pl1 && pl2) {
    
            curVal = pl1->val + pl2->val + flag;
            ListNode* tmp = new ListNode(curVal % 10);
            curNode->next = tmp;
            curNode = tmp;
            flag = 0;
            if (curVal >= 10) flag = 1;
            pl1 = pl1->next;
            pl2 = pl2->next;
        }
        while (pl1) {
    
            curVal = pl1->val + flag;
            ListNode* tmp = new ListNode(curVal % 10);
            curNode->next = tmp;
            curNode = tmp;
            flag = 0;
            if (curVal >= 10) flag = 1;
            pl1 = pl1->next;            
        }
        while (pl2) {
    
            curVal = pl2->val + flag;
            ListNode* tmp = new ListNode(curVal % 10);
            curNode->next = tmp;
            curNode = tmp;
            flag = 0;
            if (curVal >= 10) flag = 1;
            pl2 = pl2->next;            
        }
        if (flag) {
    
            ListNode* tmp = new ListNode(1);
            curNode->next = tmp;            
        }
        return ans->next;
    }
};

20. 有效的括号

代码实现（自解）

class Solution {
    
private:
    bool match(char l, char r) {
    
        if (l == '(') return r == ')';
        if (l == '{') return r == '}';
        if (l == '[') return r == ']';
        return false;
    }
public:
    bool isValid(string s) {
    
        set<char> left = {
     '(', '{', '[' };
        set<char> right = {
     ')', '}', ']'};
        stack<char> _stack;
        for (char c : s) {
    
            if (left.count(c)) {
    
                _stack.push(c);
            }
            else {
    
                if (_stack.empty()) return false;
                if (match(_stack.top(), c)) _stack.pop();
                else return false;
            }
        }
        if (!_stack.empty()) return false;
        return true;
    }
};