Java 实现 classSolution{publicbooleanisValid(String s){ Stack<Character> stack =newStack<>();for(inti=0; i<s.length(); i++) {charc=s.charAt(i);if(c =='('|| c =='['|| c =='{') { stack.push(c); }else{if(stack.empty()) {returnfalse; }chartopChar=stack.pop();if(c...
运行 AI代码解释 classSolution{protected:boolcheck(char a,char b){cout<<"a: "<<a<<"b: "<<b<<endl;bool flag=false;if(a=='('){if(b==')')flag=true;}elseif(a=='['){if(b==']')flag=true;}else{if(b=='}')flag=true;}returnflag;}public:boolisValid(string s){string str=...
*/ class Solution { public: bool isValid(string s) { if(s.size() == 0) retu...Longest Valid Parentheses Longest Valid Parentheses question: 用Stack的方法,一个右括号能消去Stack顶上的一个左括号。为了能够计算括号对的长度我们还需要记录括号们的下标。这样在弹出一个左括号后,我们可以根据当前...
这里还有一种情况,就是stack中还有残留的字符没有得到匹配,即此时stack不是空的,这时说明S不是valid的,因为只要valid,一 定全都可以得到匹配使左括号弹出。 publicclassSolution {publicbooleanisValid(String s) { Stack<Character> stack=newStack<Character>();for(inti=0 ; i<s.length(); i++){charc=s...
determine if the input string is valid. The brackets must close in the correct order,"()"and"()[]{}"are all valid but"(]"and"([)]"are not. 原题链接:https://oj.leetcode.com/problems/valid-parentheses/ 题目:给定一个仅包括'(',')','{','}','['和']' 的字符串,检測输入的串是...
class Solution { private: static bool IsPair(char c1, char c2) { if (c1 == '(' && c2 == ')' || c1 == '{' && c2 == '}' || c1 == '[' && c2 == ']') { return true; } return false; } public: bool isValid(string s) { ...
classSolution{public:boolisValid(string s){// Start typing your C/C++ solution below// DO NOT write int main() functionstack<char>st;for(int i=0;i<s.size();i++){if(s[i]=='('||s[i]=='{'||s[i]=='['){st.push(s[i]);}if(s[i]==')'){if(st.empty()||st.top()...
=m[i]:## 如果栈里面抽出最上层的元素,不是任意左括号。比如一开始遍历的第一个字符是右括号,那边 !=A, 则是非法字符returnFalse## 则是无效字符串else:## 如果是左括号stack.append(i)## 就压入栈内returnlen(stack)==1## 看栈是否为空。==1 表示有A在栈底,!= 1 就表示Solution().isValid("...
Leetcode上Valid Parentheses问题的时间复杂度是多少? 题目: 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 “...
class Solution { boolean isValidParentheses(char begin, char end) { if (('{' == begin && '}' == end) || ('[' == begin && ']' == end) || ('(' == begin && ')' == end)){ return true; } return false; } public boolean isValid(String s) { if (null == s || "...