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.
方法一、找到第一个右括号，然后和左边的字符一起借助哈希表映射，判断是否成对。
1 public static boolean isValid(String s) { 2 if(s.length()<2 || s.length()%2==1) 3 return false; 4 Map<Character,Integer> map=new HashMap<Character,Integer>(); 5 map.put('(', 1); 6 map.put(')', 9); 7 map.put('{', 2); 8 map.put('}', 8); 9 map.put('[', 3); 10 map.put(']', 7); 11 boolean result=true; 12 for(int i=0;i<s.length();i++) 13 { 14 if(map.get(s.charAt(i))<5) 15 { 16 if(i==s.length()-1)//防止越界，也没找到，最后一个肯定不是右括号 17 return false; 18 continue; 19 } 20 if(i-1<0)//排除出现}{的情况 21 return false; 22 if(map.get(s.charAt(i-1))+map.get(s.charAt(i))==10) 23 { 24 s=s.substring(0, i-1)+(i+1<s.length()?s.substring(i+1, s.length()):""); 25 if(s.length()!=0) 26 { 27 result=isValid(s); 28 } 29 break; 30 } 31 else 32 { 33 result=false; 34 } 35 36 } 37 return result; 38 }
方法二，栈
public boolean isValid(String s) { if(s.length()<1 || s.length()%2!=0) return false; Stack<Character> stack=new Stack<Character>(); for(int i=0;i<s.length();i++) { if(s.charAt(i)=='{' || s.charAt(i)=='[' || s.charAt(i)=='(') { stack.push(s.charAt(i)); if(i==s.length()-1)//最后一个 return false; continue; } switch (s.charAt(i)) { case '}': if(stack.size()==0) return false; Character c1=stack.pop(); if(c1!='{') { return false; } break; case ')': if(stack.size()==0) return false; Character c2=stack.pop(); if(c2!='(') { return false; } break; case ']': if(stack.size()==0) return false; Character c3=stack.pop(); if(c3!='[') { return false; } break; } } return true; }