描述
Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
An input string is valid if:
- Open brackets must be closed by the same type of brackets.
- Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.
Example 1:
Input: "()"
Output: true
複製程式碼Example 2:
Input: "()[]{}"
Output: true
複製程式碼Example 3:
Input: "(]"
Output: false
複製程式碼Example 4:
Input: "([)]"
Output: false
複製程式碼Example 5:
Input: "{[]}"
Output: true
複製程式碼思路
使用棧的資料結構，當遇到左括號時入棧，碰到匹配的右括號是出棧。
要注意棧空時的出棧溢位和不匹配的右括號的報錯。
class Solution {
public boolean isValid(String s) {
LinkedList<Character> stack = new LinkedList<Character>();
int len = s.length(), count = 0;
//對字串進行遍歷
while(count < len){
if(s.charAt(count) == '('||s.charAt(count) == '['||s.charAt(count) == '{'){
//左括號入棧
stack.add(s.charAt(count));
}else{
//右括號出棧
//當棧空/未匹配到時返回false，否則按規則匹配
if(stack.size() == 0){
return false;
}else if(s.charAt(count) == ')' && stack.getLast() == '('){
stack.removeLast();
}else if(s.charAt(count) == ']' && stack.getLast() == '['){
stack.removeLast();
}else if(s.charAt(count) == '}' && stack.getLast() == '{'){
stack.removeLast();
}else{
return false;
}
}
count++;
}
//遍歷整個輸入後棧空即為合法
if(stack.size() == 0){
return true;
}else{
return false;
}
}
}複製程式碼Runtime: 1 ms, faster than 98.70% of Java online submissions for Valid Parentheses.
Memory Usage: 37.4 MB, less than 5.06% of Java online submissions for Valid Parentheses.
這裡使用的是LinkedList作為儲存結構，其實java中自帶是stack結構的。