【LeetCode】- Valid Parentheses（有效括弧）

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[ 問題: ]

Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine if the input string is valid.
直譯：給定一個字串，該串包含字元‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘, ‘]‘， 請判斷它是不是有效

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

[ 解法: ]

解題思路：使用堆棧資料結構， 進入的字元如果不能左右匹配就壓棧，如果可以匹配就彈棧，並開始下一次迴圈。

當迴圈結束後，如果堆棧仍為空白的話，說明是有效括弧。

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 mapMap<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 loopif (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 emptyif (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;}}

【LeetCode】- Valid Parentheses（有效括弧）