20 Valid Parentheses
水題,能想到stack就能做
bool isValid(string s) {
stack<char>scheck;
for (int i = 0; i < s.size(); ++i) {
if (s[i] == '(' || s[i] == '[' || s[i] == '{'){
if (s[i] == '(')scheck.push(')');
else if (s[i] == '{')scheck.push('}');
else scheck.push(']');
}
else {
if (scheck. empty())return false;
if (s[i] == scheck.top()) {
scheck.pop();
}
else return false;
}
}
if (!scheck.empty())return false;
return true;
}
相關推薦
#20 Valid Parentheses
ica ini content ng- height parent sta clas cas 題目鏈接:https://leetcode.com/problems/valid-parentheses/ Given a string containi
20. Valid Parentheses
emp val pre rect nbsp not blog java case Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine if the
【leetcode】20. Valid Parentheses
lines fadein cti return nes string data- 復雜度 ack @requires_authorization @author johnsondu @create_time 2015.7.13 11:03 @url [v
LeetCode記錄之20——Valid Parentheses
{} 循環 i++ nth return urn term ket style 本題主要是找是否有匹配的字符串,因為還沒有復習到棧之類的知識點,只能還是采用暴力方法了,後期會補上更加優化的算法。我的思路就是先遍歷一遍找是否有匹配的符號,有的話就刪除,然後繼續遍歷,直至結束
leetcode練習:20. Valid Parentheses
nsh true ets str valid ack span brackets || Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine if t
20. Valid Parentheses括號匹配
ase action style spa break ive clas not str 20 Valid Parentheses Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘
LeetCode:20. Valid Parentheses(Easy)
實現 16px parent 數字 字符串 java 給定 遇到 ray 1. 原題鏈接 https://leetcode.com/problems/valid-parentheses/description/ 2. 題目要求 給定一個字符串s,s只包含‘(‘, ‘)‘,
20. Valid Parentheses(括號匹配,用桟)
char val solution 如果 close nth fix body -c Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine if
20. Valid Parentheses檢驗括號字符串的有效性
輸入 情況 brackets har 不存在 close ring HA IT [抄題]: Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine i
LeetCode - 20. Valid Parentheses(0ms)
string out etc contain not The leetcode for side Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine
LeetCode 20. Valid Parentheses
true shm bracket .get 匹配 brush imp charat nta Given a string containing just the characters ‘(‘, ‘)‘, ‘{‘, ‘}‘, ‘[‘ and ‘]‘, determine if
#Leetcode# 20.Valid Parentheses
https://leetcode.com/problems/valid-parentheses/description/ Given a string containing just the characters '(', ')', '{', '}',
20. Valid Parentheses 有效的括號
題目 程式碼部分一(8ms 86.26%) class Solution { public boolean isValid(String s) { Stack<Character> st = new S
[leetcode] 20. Valid Parentheses (easy)
原題連結 匹配括號 思路: 用棧,遍歷過程中,匹配的成對出棧;結束後,棧空則對,棧非空則錯。 Runtime: 4 ms, faster than 99.94% of Java class Solution { public boolean isValid(String s) {
【演算法設計與分析作業題】第十一週:20. Valid Parentheses
題目 C++ solution class Solution { public: bool isValid(string s) { stack<char> cstack; for (int i = 0; i < s.si
【LeetCode】20 Valid Parentheses 有效括號
給定一個只包括 ‘(’,’)’,’{’,’}’,’[’,’]’ 的字串,判斷字串是否有效。 有效字串需滿足: 左括號必須用相同型別的右括號閉合。 左括號必須以正確的順序閉合。 注意空字串可被認為是有效字串。 示例 1: 輸入: “()” 輸出: true 示例 2: 輸入: “()[
leetcode-20 valid-parentheses(有效的括號)
先看一下題目描述: 通過題目描述可以清楚的明白題目規則,輸出true或者false。這道題需要用借用棧來實現,不說理解直接放程式碼 1 public boolean isValidParentheses(String s) { 2 // Write your code her
【LeetCode】20. Valid Parentheses - Java實現
文章目錄 1. 題目描述: 2. 思路分析: 3. Java程式碼: 1. 題目描述: Given a string containing just the characters '(', ')', '{', '}', '[' an
20 Valid Parentheses
水題,能想到stack就能做 bool isValid(string s) { stack<char>scheck; for (int i = 0; i < s.size(); ++i) { if (s[i] == '(' || s[i] == '[' || s
[leetcode]20. Valid Parentheses
這個題果真很easy 但是注意“【”這種這有左邊的情況 class Solution { public boolean isValid(String s) { if(s.length()==0||s==null)return true;