【LeetCode-栈】有效的括号
题目来源于 LeetCode 上第 20号（Valid Parentheses）问题：有效的括号。题目难度为 Easy。面试
题目地址：https://leetcode.com/problems/valid-parentheses/数组
题目描述
Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.bash
给定一个字符串, 只包括 '('，')'，'{'，'}'，'['，']'，判断字符串是否有效ide
An input string is valid if:post
有效字符串须要知足如下条件：ui
-
Open brackets must be closed by the same type of brackets. 左括号必须用相同类型的右括号闭合spa
-
Open brackets must be closed in the correct order. 左括号必须以正确的顺序闭合3d
Note that an empty string is also considered valid.code
注意空字符串可被认为是有效字符串。cdn
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
复制代码
题目解析
- 遍历字符串
- 遇到左括号，则将其压入栈中
- 若是遇到右括号：
- 当前栈为空，直接返回false;
- 当前右括号对应的左括号，与栈顶元素不相等，直接返回false
- 循环结束以后，判断栈是否为空，不为空返回false
附上提交结果：
代码实现
class Solution {
public boolean isValid(String s) {
if (s == null) return false;
Stack<Character> stack = new Stack();
//遍历字符串
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
//遇到左括号，则将其压入栈中
if (c == '(' || c == '{' || c == '[') {
stack.push(c);
} else {
//当前栈为空，直接返回false
if (stack.isEmpty()) return false;
char top = stack.pop();
char bracket = '0';
if (c == ')') {
bracket = '(';
} else if (c == '}') {
bracket = '{';
} else if (c == ']') {
bracket = '[';
}
//当前右括号对应的左括号，与栈顶元素不相等，直接返回false
if (top != bracket) {
return false;
}
}
}
return stack.isEmpty();
}
}
复制代码