[LeetCode] Valid Phone Numbers 验证电话号码 Given a text filefile.txtthat contains list of phone numbers (one per line), write a one liner bash script to print all valid phone numbers. You may assume that a valid phon
(123) 456-7890 Your script should output the following valid phone numbers: 987-123-4567 (123) 456-7890 解答 https://leetcode.com/problems/valid-phone-numbers/discuss/55478/Grep-e-solution-with-detailed-explanation-good-for-those-new-to-regex 1grep-e'\(^[0-9]\{3\}-[0-9]\{3\}-[...
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition. class States(object): def __init__(self): self.init = 0 self.decimal = 1 self.decpoint = 2 self...
按上面的规则便行。 #include <iostream>usingnamespacestd;classSolution {public:boolisNumber(constchar*s) {intidx =0;for(;s[idx]=='';idx++);if(s[idx]=='-'||s[idx]=='+') idx++;intPoint=0,Num=0;for(;(s[idx]>='0'&&s[idx]<='9')||s[idx]=='.';idx++) s[idx]=='....
Valid Parentheses - LeetCode 注意点 考虑输入为空的情况 解法 解法一:如果是'('、'{'、'['这三者就入栈,否则就判断栈是否为空和栈顶括号是否与之匹配。注意两个判断顺序不可以颠倒,不然会runtime error。时间复杂度为O(n) classSolution{public:boolisValid(string s){ ...
classSolution {public:boolisValid(strings) { stack<char>stk;for(inti =0; i < s.size(); i ++) {switch(s[i]) {case'(': stk.push(s[i]);break;case')':if(!stk.empty() && stk.top() =='(') { stk.pop();break; }elsereturnfalse;case'[': ...
classSolution { public: intlongestValidParentheses(string s) { constintlen = s.size(); boolisValid[len][len]; memset(isValid, 0,sizeof(isValid)); intres = 0; for(inti = 0; i < len-1; i++) if(s[i] =='('&& s[i+1] ==')') ...
[leetcode]Valid Parentheses @ Python 原题地址:https://oj.leetcode.com/problems/valid-parentheses/ 题意: Given a string containing just the characters'(',')','{','}','['and']', determine if the input string is valid. The brackets must close in the correct order,"()"and"()[]{}...
https://oj.leetcode.com/problems/valid-parentheses/ 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. ...
};classSolution {public: ListNode* removeNthFromEnd(ListNode* head,intn) {intcnt=0;//not harm head nodeListNode* node=head;while(node) {++cnt; node=node->next; }intindex=cnt-n-1;if(0==index) { head=head->next; }else{ node=head;while(--index) ...