【LeetCode】#20有效的括号(Valid Parentheses)
题目描述
给定一个只包括 ‘(’，’)’，’{’，’}’，’[’，’]’ 的字符串，判断字符串是否有效。
有效字符串需满足：
1.左括号必须用相同类型的右括号闭合。
2.左括号必须以正确的顺序闭合。
注意空字符串可被认为是有效字符串。
示例
示例 1:
输入: “()”
输出: true
示例 2:
输入: “()[]{}”
输出: true
示例 3:
输入: “(]”
输出: false
示例 4:
输入: “([)]”
输出: false
示例 5:
输入: “{[]}”
输出: true
Description
Given a string containing just the characters ‘(’, ‘)’, ‘{’, ‘}’, ‘[’ and ‘]’, determine if the input string is valid.
An input string is valid if:
1.Open brackets must be closed by the same type of brackets.
2.Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.
Example
Example 1:
Input: “()”
Output: true
Example 2:
Input: “()[]{}”
Output: true
Example 3:
Input: “(]”
Output: false
Example 4:
Input: “([)]”
Output: false
Example 5:
Input: “{[]}”
Output: true
解法
class Solution {
public boolean isValid(String s) {
char[] ch = s.toCharArray();
int len = ch.length;
Stack<Character> stack = new Stack<>();
if(s.length()%2!=0)
return false;
if(s.length()==0)
return true;
if(ch[0]==')' || ch[0]=='}' || ch[0]==']')
return false;
stack.push(ch[0]);
for(int i=1; i<len; i++){
switch(ch[i]){
case '(':
stack.push(ch[i]);
break;
case '[':
stack.push(ch[i]);
break;
case '{':
stack.push(ch[i]);
break;
case ')':
if(stack.peek()=='(')
stack.pop();
else
return false;
break;
case ']':
if(stack.peek()=='[')
stack.pop();
else
return false;
break;
case '}':
if(stack.peek()=='{')
stack.pop();
else
return false;
break;
default:
break;
}
}
if(stack.isEmpty())
return true;
return false;
}
}