• 括号问题


    20. Valid Parentheses(括号匹配,用桟)

    class Solution {
    public:
        bool isValid(string s) {
            stack<char> stk;
            for (char c : s) {
                if (c == '(' || c == '[' || c=='{') {
                    stk.push(c);
                }  else if (!stk.empty() && leftof(c)==stk.top()) {
                    stk.pop();
                } else {
                    return false;
                }
            }
            return stk.empty();
        }
        char leftof(char c) {
            if (c == ')') return '(';
            if (c == '}') return '{';
            if (c == ']') return '[';
            return c;
        }
    };

    22. Generate Parentheses(回溯)

    class Solution {
    public:
        vector<string> res;
            bool is_vilid(const string& str) {
            int balance = 0;
            for (char c : str) {
                if (c == '(') {
                    ++balance;
                } else {
                    --balance;
                }
                if (balance < 0) {
                    return false;
                }
            }
            return balance == 0;
        }
    
        void dfs(string& path,int level,int n) {
             if (level>n ) { 
                 return;
             }    
            if (level==n ) {
                if (is_vilid(path)) res.emplace_back(path);
                return;
            }
            string path_back = path;
            path.append("(");
            dfs(path,level+1,n);
            path.pop_back();
    
            path.append(")");
            dfs(path,level+1,n);
            path.pop_back();
            
        }
        vector<string> generateParenthesis(int n) {
            string path = "";
            dfs(path,0,n*2);
            return res;
        }
    };



    32. Longest Valid Parentheses(最长括号匹配,hard)

    class Solution {
    public:
        int longestValidParentheses(string s) {
            int res = 0;
            int l_cnt =0;
            int r_cnt = 0;
            for(int i = 0; i < s.size();i++) {
                if (s[i] == '(') l_cnt++;
                else r_cnt++;
                if (l_cnt==r_cnt) res = max(res,r_cnt*2);
                if (l_cnt<r_cnt) l_cnt = r_cnt = 0;
        
            }
            r_cnt = l_cnt = 0;
            for(int i = s.size()-1; i >= 0;i--) {
                if (s[i] == '(') l_cnt++;
                else r_cnt++;
                if (l_cnt==r_cnt) res = max(res,r_cnt*2);
                if (r_cnt<l_cnt) l_cnt = r_cnt = 0;   
            }
            return res;
        }
    };

    301. 删除无效的括号(dfs)

    class Solution {
    public:
        set<string> res;
        void dfs(string path,int level,string& s,int l_cnt, int r_cnt,int min_cnt) {
            if (l_cnt < r_cnt || l_cnt> min_cnt || r_cnt>min_cnt || level>s.size()) return;
            if (level == s.size()) {
                if (l_cnt==r_cnt && l_cnt == min_cnt) {
                    res.insert(path);
                }
            }
            char c = s[level];
            if (c =='(') {
                dfs(path+c,level+1,s,l_cnt+1,r_cnt,min_cnt);
                dfs(path,level+1,s,l_cnt,r_cnt,min_cnt);
            } else if (c == ')') {
                dfs(path+c,level+1,s,l_cnt,r_cnt+1,min_cnt);
                dfs(path,level+1,s,l_cnt,r_cnt,min_cnt);
            } else {
                dfs(path+c,level+1,s,l_cnt,r_cnt,min_cnt);
            }
        }
        vector<string> removeInvalidParentheses(string s) {
            int l_cnt = 0;
            int r_cnt = 0;
            for(auto c: s) {
                if (c=='(') l_cnt++;
                else if (c==')') { // )( 避免这种case
                    if (l_cnt > r_cnt) {
                    r_cnt++;
                    }
                }
            }
            int min_cnt = min(l_cnt,r_cnt);
            dfs("",0,s,0,0,min_cnt);
            vector<string> out;
            for(auto i : res) {
                out.emplace_back(i);
            }
            return out;
        }
    };

    678. 有效的括号字符串

    class Solution {
    public:
        bool checkValidString(string s) {
            int s_cnt = 0;
            int l_cnt =0;
            int r_cnt = 0;
            for(int i = 0; i < s.size();i++) {
                if (s[i] == '(') l_cnt++;
                else if (s[i]==')') r_cnt++;
                else s_cnt++;
                if (l_cnt + s_cnt <r_cnt) return false;
            }
            s_cnt = r_cnt = l_cnt = 0;
            for(int i = s.size()-1; i >= 0;i--) {
                if (s[i] == '(') l_cnt++;
                else if (s[i]==')') r_cnt++;
                else s_cnt++;
                if (r_cnt + s_cnt <l_cnt) return false;
            }
            return true;;
        }
    };

    921. 使括号有效的最少添加

    class Solution {
    public:
        int minAddToMakeValid(string s) {
            int l_cnt =0;
            int r_cnt = 0;
            int res = 0;
            for(int i = 0; i < s.size();i++) {
                if (s[i] == '(') l_cnt++;
                else  r_cnt++;
                if (r_cnt>l_cnt) {
                    res++;
                    l_cnt++;
                }
            }        
            r_cnt = l_cnt = 0;
            for(int i = s.size()-1; i >= 0;i--) {
                if (s[i] == '(') l_cnt++;
                else r_cnt++;
                if (l_cnt>r_cnt) {
                    res++;
                    r_cnt++;
                }
            }
            return res;
        }
    };
  • 相关阅读:
    In Java, how do I read/convert an InputStream to a String? Stack Overflow
    IFrame自动更改大小
    [置顶] 获取服务器上格式为JSON和XML两种格式的信息的小程序
    Qt VS MFC
    [技术分享]使用 UAG 发布 RemoteAPP
    linux2.6.32在mini2440开发板上移植(11)之配置USB外设
    MFC控件(2):Text_Edit_Control
    CentOS 6.4 安装 Oracle 10g2 备记
    sql lite 二个数据库之间的表进行复制
    变量和函数的定义和声明
  • 原文地址:https://www.cnblogs.com/zle1992/p/15971487.html
Copyright © 2020-2023  润新知