Leetcode 20.Valid Parenthese

我的思路

很简单, 遇到 ()、[]、{} 这三对符号的左边的就入栈, 遇到右边的就直接将栈顶元素与其对应进行比较, 能组成一对符号就出栈, 否则就返回 false. 因为有 6 种情况就使用 switch 语句 :

import java.util.Stack;
class Solution {public boolean isValid(String s) {Stack<Character> cstack = new Stack<>();char[] sc = s.toCharArray();for(char c : sc) {switch(c) {case '(' :cstack.push(c);break;case '[' :cstack.push(c);break;case '{' :cstack.push(c);break;case ')' :if (!cstack.isEmpty() && cstack.peek().equals('(')) {cstack.pop();} else {return false;}break;case ']' :if (!cstack.isEmpty() && cstack.peek().equals('[')) {cstack.pop();} else {return false;}break;case '}' :if (!cstack.isEmpty() && cstack.peek().equals('{')) {cstack.pop();} else {return false;}break;default :return false;}}return cstack.isEmpty();}
}

大神的思路

代码简洁了很多, 首先是字符串转换直接在 for 语句中进行, 然后遇到符号中左边的字符, 直接将符号对应的右边字符入栈, 这样可以直接比较. 学习了.

class Solution {public boolean isValid(String s) {Stack<Character> stack = new Stack<>();for (char c : s.toCharArray()) {if (c == '(' {stack.push(')');} else if (c == '[') {stack.push(']');} else if (c == '{') {stack.push('}');} else if (stack.isEmpty() || stack.pop() != c) return false;}return stack.isEmpty();}
}

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Leetcode 20.Valid Parenthese

我的思路

大神的思路

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