【LeetCode】- Valid Parentheses（有效的括号）

[ 问题: ]

Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine if the input string is valid.

直译：给定一个字符串，该串包含字符‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘, ‘]‘， 请判断它是不是有效的

The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.

[ 解法: ]

解题思路：使用堆栈数据结构， 进入的字符如果不能左右匹配就压栈，如果可以匹配就弹栈，并开始下一次循环。

当循环结束后，如果堆栈仍为空的话，说明是有效的括号。

import java.util.HashMap;
import java.util.Map;
import java.util.Stack;

public class Solution {
	public static void main(String[] args) {
		System.out.println(new Solution().isValid("{[()]}"));
	}

	public boolean isValid(String s) {
		if (s != null && !"".equals(s.trim())) {
			// init the character map
			Map<Character, Character> map = init();
			Stack<Character> stack = new Stack<Character>();

			char[] ch = s.toCharArray();
			for (int i = 0; i < ch.length; i++) {
				// the first time come into the loop
				if (stack.size() == 0) {
					stack.push(ch[i]);
				} else {
					Character pre = stack.peek();
					if (pre.equals(map.get(ch[i]))) {
						stack.pop();
						continue;
					}
					stack.push(ch[i]);
				}
			}

			// the stack is empty
			if (stack.size() == 0) {
				return true;
			}
		}
		return false;
	}

	/**
	 * data init.
	 * @return Map<Character>
	 */
	public Map<Character, Character> init() {
		Map<Character, Character> map = new HashMap<Character, Character>();
		map.put('(', ')');
		map.put(')', '(');
		map.put('{', '}');
		map.put('}', '{');
		map.put('[', ']');
		map.put(']', '[');
		return map;
	}
}