public String removeOuterParentheses(String S) { StringBuilder res = new StringBuilder(); LinkedList<Character> stack = new LinkedList<>(); for (int st = 0, ed = 0; ed != S.length();) { st = ed; stack.addLast(S.charAt(ed)); ed++; while (stack.size() != 0 && ed != S.le...
publicclassRemoveOutermostParentheses {/*解法一:栈,( 进栈,) 出栈,栈为空时,则找到原语,然后去括号。*/publicString removeOuterParentheses(String S) { StringBuilder sb=newStringBuilder(); Stack<Character> stack=newStack<>();intstart=0;intend=0;for(inti=0;i<S.length();i++){charc=S.charAt(...
【leetcode_easy_stack】1021. Remove Outermost Parentheses problem 1021. Remove Outermost Parentheses solution#1: stack; code solution#2: code 参考 1.leetcode_easy_stack_1021. Remove Outermost Parentheses; 2.【leetcode刷题】1021. Remove Outermost Parentheses; 完 各美其美,美美与共,不和他人作比较,...
Given a valid parentheses string S, consider its primitive decomposition: S = P_1 + P_2 + ... + P_k, where P_i are primitive valid parentheses strings. Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S. Example 1: Input: "...
在S中,左括号"("和右括号")"是成对出现的,即在每一个原始有效字符串中,左括号和右括号的数量是相同的。 思路:从左到右遍历S中的字符,定义两个变量left和right,记录遇到的左括号、右括号的数量,当左括号和右括号的数量相等时,说明遇到了一个原始有效字符串,去掉头尾的括号,借助字符串截取完成,拼接到StringBui...
Given a valid parentheses string S, consider its primitive decomposition: S = P_1 + P_2 + ... + P_k, where P_i are primitive valid parentheses strings. Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S. Example 1: Input: "...
具体看code。 Java Solution: Runtime: 2 ms, faster than 97.83% Memory Usage: 38.4 MB, less than 5.19% 完成日期:02/15/2020 关键点:stack classSolution {publicString removeOuterParentheses(String S) { StringBuilder s=newStringBuilder();intopened = 0;for(charc : S.toCharArray()) {if(c ==...
网址:https://leetcode.com/problems/remove-outermost-parentheses/ 使用栈的思想,选择合适的判断时机 classSolution {public:stringremoveOuterParentheses(stringS) { stack<char>s_ch;stringans; stack<char>temp;intl =0, r =0;for(inti =0; i<S.size(); i++) ...
Leetcode 1021. Remove Outermost Parentheses 括号匹配想到用栈来做: classSolution:defremoveOuterParentheses(self, S: str) ->str: size=len(S)ifsize==0:return""i=0 strings=[]whilei<size: stack=[S[i]] string=S[i] i+=1whilei<size:ifS[i]=='(':...
1classSolution2{3public:4stringremoveOuterParentheses(stringS)5{6stringrnt;7stack<char>s;8s.push(S[0]);910for(inti =1; i < S.size();i ++)11{12if(S[i]=='(')13{14s.push(S[i]);15rnt +='(';16}17elseif(s.size()>1&& S[i]==')')18{19s.pop();20rnt +=')';21}22...