【LeetCode-面試演算法經典-Java實現】【020-Valid Parentheses(括弧驗證)】,-javaparentheses

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【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。

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