[LeetCode] 422. Valid Word Square_Easy】的更多相关文章

Given a sequence of words, check whether it forms a valid word square. A sequence of words forms a valid word square if the kth row and column read the exact same string, where 0 ≤ k < max(numRows, numColumns). Note: The number of words given is at l…
原题链接在这里:https://leetcode.com/problems/valid-word-square/ 题目: Given a sequence of words, check whether it forms a valid word square. A sequence of words forms a valid word square if the kth row and column read the exact same string, where 0 ≤ k < max(…
作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 拼接出每一列的字符串 日期 题目地址:https://leetcode-cn.com/problems/valid-word-square/ 题目描述 Given a sequence of words, check whether it forms a valid word square. A sequence of words forms a va…
Given a non-empty string s and an abbreviation abbr, return whether the string matches with the given abbreviation. A string such as "word" contains only the following valid abbreviations: ["word", "1ord", "w1rd", &…
似乎可以沿着对角线往右往下检查,也可以正常按题设检查. 我用的后者.. public class Solution { public boolean validWordSquare(List<String> words) { if(words.size() <= 1) return true; if(words.size() != words.get(0).length()) return false; for(int i = 0; i < words.get(0).length(…
Given a positive integer num, write a function which returns True if num is a perfect square else False. Note: Do not use any built-in library function such as sqrt. Example 1: Input: 16 Returns: True Example 2: Input: 14 Returns: False Use Newton Me…
Given a sequence of words, check whether it forms a valid word square. A sequence of words forms a valid word square if the kth row and column read the exact same string, where 0 ≤k < max(numRows, numColumns). Note: The number of words given is at le…
Given a sequence of words, check whether it forms a valid word square. A sequence of words forms a valid word square if the kth row and column read the exact same string, where 0 ≤ k < max(numRows, numColumns). Note: The number of words given is at l…
A string such as "word" contains the following abbreviations: ["word", "1ord", "w1rd", "wo1d", "wor1", "2rd", "w2d", "wo2", "1o1d", "1or1", "…
第一种方法,用栈实现,最容易想到,也比较容易实现,每次碰到‘)’时update max_len,由于要保存之前的‘(’的index,所以space complexity 是O(n) // 使用栈,时间复杂度 O(n),空间复杂度 O(n) class Solution { public: int longestValidParentheses(string s) { , last = -; stack<int> lefts; ; i < s.size(); ++i) { if (s[i]…