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.
没考虑到的情况有：input: [, expected: false; 语法上也犯了错误： 我定义的stack的泛型为char，泛型不能为primitive datatype,
Primitive types such as char cannot be used as type parameters in Java. You need to use the wrapper type:
Stack<Character> stack =newStack<Character>(); 建议有时间再多读读泛型部分
stack做法：（默认方法，用switch）
1 public boolean isValid(String s) { 2 if(s==null || s.length()==0) 3 return true; 4 LinkedList<Character> stack = new LinkedList<Character>(); 5 for(int i=0;i<s.length();i++) 6 { 7 switch(s.charAt(i)) 8 { 9 case '(': 10 case '{': 11 case '[': 12 stack.push(s.charAt(i)); 13 break; 14 case ')': 15 if(stack.isEmpty() || stack.pop()!='(') 16 return false; 17 break; 18 case '}': 19 if(stack.isEmpty() || stack.pop()!='{') 20 return false; 21 break; 22 case ']': 23 if(stack.isEmpty() || stack.pop()!='[') 24 return false; 25 break; 26 default: 27 break; 28 } 29 } 30 if(stack.isEmpty()) 31 return true; 32 return false; 33 }
用数组模拟stack, 用一个head指示栈头的下一个:
1 public class Solution { 2 public boolean isValid(String s) { 3 char[] stack = new char[s.length()]; 4 int head = 0; 5 for(char c : s.toCharArray()) { 6 switch(c) { 7 case '{': 8 case '[': 9 case '(': 10 stack[head++] = c; 11 break; 12 case '}': 13 if(head == 0 || stack[--head] != '{') return false; 14 break; 15 case ')': 16 if(head == 0 || stack[--head] != '(') return false; 17 break; 18 case ']': 19 if(head == 0 || stack[--head] != '[') return false; 20 break; 21 } 22 } 23 return head == 0; 24 25 } 26 }