内容简介: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.
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
难度:easy
题目:
给定仅包含字符'(', ')', '{', '}', '[' 和 ']'的字符串, 判定这个字符串是否有效。
输入字符串是否有效:
开括号必需与同类型括号关闭。
开括号必需要以正确的顺序关闭。
注意 空字符串被认为是有效的字符串。
思路:stack
Runtime: 5 ms, faster than 74.68% of Java online submissions for Valid Parentheses.
Memory Usage: 26.3 MB, less than 52.20% of Java online submissions for Valid Parentheses.
class Solution { public boolean isValid(String s) { // stack Stack<Character> stack = new Stack<>(); for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); if (c == '(') { stack.push(')'); } else if (c == '{') { stack.push('}'); } else if (c == '[') { stack.push(']'); } else { if (stack.isEmpty() || stack.pop() != c) { return false; } } } return stack.isEmpty(); } }
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