这是一道简单的题目,方法为利用栈针对数据的先入后出的特性进行解题。单从题目看,输入成立的可能总共有三种,分别为 :
- “()[]{}”,两两相结合,中间不含有其他类型括号
- “{[]}”,形成一个对称的形式,从中间区分左边和右边分别为括号的左边和右边。
- 前两种进行结合。
对于此题栈的方法毫无疑问最为高效,复杂度为n,将左括号和右括号对应关系变为哈希表,键值对。针对输入进行遍历,左括号放入哈希表中,一旦出现右括号则进行判断,如果与栈中的top元素相对应,则将栈中元素抛出,接着进行判断,否则直接返回false。
代码如下:
class Solution {
// Hash table that takes care of the mappings.
private HashMap<Character, Character> mappings;
// Initialize hash map with mappings. This simply makes the code easier to read.
public Solution() {
this.mappings = new HashMap<Character, Character>();
this.mappings.put(')', '(');
this.mappings.put('}', '{');
this.mappings.put(']', '[');
}
public boolean isValid(String s) {
// Initialize a stack to be used in the algorithm.
Stack<Character> stack = new Stack<Character>();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
// If the current character is a closing bracket.
if (this.mappings.containsKey(c)) {
// Get the top element of the stack. If the stack is empty, set a dummy value of '#'
char topElement = stack.empty() ? '#' : stack.pop();
// If the mapping for this bracket doesn't match the stack's top element, return false.
if (topElement != this.mappings.get(c)) {
return false;
}
} else {
// If it was an opening bracket, push to the stack.
stack.push(c);
}
}
// If the stack still contains elements, then it is an invalid expression.
return stack.isEmpty();
}
}
作者:LeetCode链接:https://leetcode-cn.com/problems/valid-parentheses/solution/you-xiao-de-gua-hao-by-leetcode/
这种做法十分巧妙,首先因为左括号必须存入栈中,同时之后会根据右括号去判断是否满足,因此将左括号设为值,将右括号设为key,之后如果发现输入为哈希表中包含的key,则去判断该key所对应value值是否跟顶端的元素相同,通过key去取值value,从而实现整个操作。