【LeetCode】20. Valid Parentheses - Java實現
阿新 • • 發佈:2018-11-24
文章目錄
1. 題目描述:
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
Example 2:
Input: "()[]{}"
Output: true
Example 3:
Input: "(]"
Output: false
Example 4:
Input: "([)]"
Output: false
Example 5:
Input: "{[]}"
Output: true
2. 思路分析:
題目的意思是給定一個包含一系列括號的字串,判斷其中的括號是否兩兩匹配。
括號匹配問題,很自然地想到用棧來處理,即遍歷字串,遇到左括號就入棧,遇到右括號,則出棧並判斷與當前的右括號是否匹配。
3. Java程式碼:
原始碼
:見我GiHub主頁
程式碼:
public static boolean isValid(String s) {
Stack<Character> stack = new Stack<>();
for (int i = 0; i < s.length(); i++) {
char ch = s.charAt(i);
// 如果是左括號,則入棧
if (ch == '(' || ch == '[' || ch == '{') {
stack.push(ch);
} else { // 如果是右括號,則比較其與棧頂元素是否配對
if (stack.isEmpty()) {
return false;
}
if (ch == ')' && stack.peek() != '(') {
return false;
}
if (ch == ']' && stack.peek() != '[') {
return false;
}
if (ch == '}' && stack.peek() != '{') {
return false;
}
stack.pop();
}
}
// 最後棧為空則表示完全匹配完畢
return stack.isEmpty();
}