【LeetCode-面試演算法經典-Java實現】【020-Valid Parentheses(括弧驗證)】,-javaparentheses
【020-Valid Parentheses(括弧驗證)】【LeetCode-面試演算法經典-Java實現】【所有題目目錄索引】原題
Given a string containing just the characters ‘(‘, ‘)’, ‘{‘, ‘}’, ‘[’ and ‘]’, determine if the input string is valid.
The brackets must close in the correct order, “()” and “()[]{}” are all valid but “(]” and “([)]” are not.
題目大意
給定一個只包含(‘, ‘)’, ‘{‘, ‘}’, ‘[’ 和‘]’的字串,驗證它是否是有效。括弧必須配對,並且要以正確的順序。
解題思路
用一個棧來對輸入的括弧串進行處理,如果是左括弧就入棧,如果是右括弧就與棧頂元素看是否組成一對括弧,組成就彈出,並且處理下一個輸入的括弧,如果不匹配就直接返回結果。
代碼實現
import java.util.*;public class Solution { public boolean isValid(String s) { Deque<Character> stack = new LinkedList<>(); int index = 0; Character top; while (index < s.length()) { Character c = s.charAt(index); switch (c) { case '(': case '[': case '{': stack.addFirst(c); break; case ')': if (stack.isEmpty()) { return false; } top = stack.getFirst(); if (top == '(') { stack.removeFirst(); } else if (top == '[' || top == '{') { return false; } else { stack.addFirst(c); } break; case ']': if (stack.isEmpty()) { return false; } top = stack.getFirst(); if (top == '[') { stack.removeFirst(); } else if (top == '(' || top == '{') { return false; } else { stack.addFirst(c); } break; case '}': if (stack.isEmpty()) { return false; } top = stack.getFirst(); if (top == '{') { stack.removeFirst(); } else if (top == '[' || top == '(') { return false; } else { stack.addFirst(c); } break; default: return false; } index++; } return stack.isEmpty(); }}
評測結果
點擊圖片,滑鼠不釋放,拖動一段位置,釋放後在新的視窗中查看完整圖片。
特別說明
歡迎轉載,轉載請註明出處【http://blog.csdn.net/derrantcm/article/details/46997247】
著作權聲明:本文為博主原創文章,未經博主允許不得轉載。