Valid Parentheses

来源：互联网发布：艾弗森对乔丹比赛数据编辑：程序博客网时间：2024/05/18 03:49

Problem: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.

最简单的解法是使用栈，'('入栈，')'时'('出栈，需要处理三种情况，存在很大的代码重复。

public static boolean isValid(String s) {Stack<Character> sk=new Stack<Character>();for(int i=0;i<s.length();i++){if(s.charAt(i)=='(')sk.push('(');if(s.charAt(i)==')'){if(!sk.empty()&&sk.peek()=='(')sk.pop();elsereturn false;}if(s.charAt(i)=='[')sk.push('[');if(s.charAt(i)==']'){if(!sk.empty()&&sk.peek()=='[')sk.pop();elsereturn false;}if(s.charAt(i)=='{')sk.push('{');if(s.charAt(i)=='}'){if(!sk.empty()&&sk.peek()=='{')sk.pop();elsereturn false;}}if(sk.empty())return true;elsereturn false;}

更简单的方法是使用HashMap存在'('') ''['']' '{''}'，再判断入栈出栈操作。

public static boolean isValid(String s) {Stack<Character> sk=new Stack<Character>();HashMap<Character,Character> hp=new HashMap<Character, Character>();hp.put('(', ')');hp.put('[', ']');hp.put('{', '}');for(int i=0;i<s.length();i++){char cur=s.charAt(i);if(hp.containsKey(cur))sk.push(cur);else if(hp.containsValue(cur)){if(!sk.empty()&&hp.get(sk.peek())==cur)sk.pop();elsereturn false;}}return sk.empty();}

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