20. Valid Parentheses括號匹配
阿新 • • 發佈:2017-11-12
ase action style spa break ive clas not str
20 Valid Parentheses
Given a string containing just the characters ‘(‘
, ‘)‘
, ‘{‘
, ‘}‘
, ‘[‘
and ‘]‘
, determine if the input string is valid.
The brackets must close in the correct order, "()"
and "()[]{}"
are all valid but "(]"
and "([)]"
are not.
自己代碼:
class Solution { public: bool isValid(strings) { stack<char> parent; for(char c:s){ if(c ==‘[‘||c == ‘(‘ || c == ‘{‘) parent.push(c); //左括號進棧 else if(parent.empty()) return false; else{ //有右括號且與棧頂匹配,則棧頂元素出棧 if(c ==‘]‘ && parent.top() == ‘[‘) parent.pop(); else if(c ==‘}‘ && parent.top() == ‘{‘) parent.pop(); else if(c ==‘)‘ && parent.top() == ‘(‘) parent.pop(); else return false; } } if(parent.empty()) return true; else return false; } };
巧妙的方法,discussion區發現:
遇到左括號,則使對應右括號進棧,例如:遇到“{”,進棧“}”。遇到“[”,進棧“]”。遇到“(”,進棧“)”。
非右括號,則看它是否與棧頂元素相等,相等即匹配。
public boolean isValid(String s) { Stack<Character> stack = new Stack<Character>(); for (char c : s.toCharArray()) { if (c == ‘(‘) //遇到左右括號,使括號進棧 stack.push(‘)‘); else if (c == ‘{‘) stack.push(‘}‘); else if (c == ‘[‘) stack.push(‘]‘); else if (stack.isEmpty() || stack.pop() != c) //遇到非右括號,若棧空,返回false。否則彈出棧頂元素,看其是否與當前元素相等,否則錯誤 return false; } return stack.isEmpty(); }
用switch語句寫:
bool isValid(string s) { stack<char> paren; for (char& c : s) { switch (c) { case ‘(‘: case ‘{‘: case ‘[‘: paren.push(c); break; case ‘)‘: if (paren.empty() || paren.top()!=‘(‘) return false; else paren.pop(); break; case ‘}‘: if (paren.empty() || paren.top()!=‘{‘) return false; else paren.pop(); break; case ‘]‘: if (paren.empty() || paren.top()!=‘[‘) return false; else paren.pop(); break; default: ; // pass } } return paren.empty() ; }
20. Valid Parentheses括號匹配