【[LeetCode] Word Break 解题思路】的更多相关文章

[LeetCode] Word Break 解题思路

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, givens = "leetcode",dict = ["leet", "code"]. Return true because &…

[LeetCode] Word Break II 解题思路

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences. For example, givens = "catsanddog",dict = ["cat", "cats"…

LeetCode: Word Break II 解题报告

Word Break II Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences. For example, given s = "catsanddog", dict = ["cat",…

【LeetCode】139. Word Break 解题报告（Python & C++）

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 日期 题目地址:https://leetcode.com/problems/unique-binary-search-trees/description/ 题目描述 Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, de…

[leetcode]Word Break II @ Python

原题地址:https://oj.leetcode.com/problems/word-break-ii/ 题意: Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences. For example, givens = "c…

[LeetCode] Word Break II 拆分词句之二

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences. For example, given s = "catsanddog", dict = ["cat", "cats&quo…

LeetCode ||& Word Break && Word Break II（转）——动态规划

一. Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, givens ="leetcode",dict =["leet", "code"]. Return true because&…

LeetCode:Word Break II（DP）

题目地址:请戳我 这一题在leetcode前面一道题word break 的基础上用数组保存前驱路径,然后在前驱路径上用DFS可以构造所有解.但是要注意的是动态规划中要去掉前一道题的一些约束条件(具体可以对比两段代码),如果不去掉则会漏掉一些解(前一道题加约束条件是为了更快的判断是字符串是够能被分词,这里是为了找出所有分词的情况) 代码如下: class Solution { public: vector<string> wordBreak(string s, unordered_set<…

LeetCode Word Break II

原题链接在这里:https://leetcode.com/problems/word-break-ii/ 题目: Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences. For example, givens = "c…

[leetcode]Word Break @ Python

原题地址:https://oj.leetcode.com/problems/word-break/ 题意: Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, givens = "leetcode",dict = [&quo…

[Leetcode] word break ii拆分词语

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences. For example, givens ="catsanddog",dict =["cat", "cats",…

[LeetCode] Maximum Gap 解题思路

Given an unsorted array, find the maximum difference between the successive elements in its sorted form. Try to solve it in linear time/space. Return 0 if the array contains less than 2 elements. You may assume all elements in the array are non-negat…

【LeetCode】Word Break 解题报告

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, given s = "leetcode", dict = ["leet", "code"]. Return true because…

LeetCode: Word Break I && II

I title: https://leetcode.com/problems/word-break/ Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, givens = "leetcode",dict = ["l…

LeetCode -- Word Break 动态规划，详细理解

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, given s = "leetcode", dict = ["leet", "code"]. Return true because…

[Leetcode] word break 拆分词语

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, givens ="leetcode",dict =["leet", "code"]. Return true because&quo…

[LeetCode] Word Break 拆分词句

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, givens = "leetcode",dict = ["leet", "code"]. Return true because &…

Leetcode Word Break

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, givens = "leetcode",dict = ["leet", "code"]. Return true because &…

LeetCode:Word Break（DP）

题目地址:http://oj.leetcode.com/problems/word-break/ 简单的动态规划问题,采用自顶向下的备忘录方法,代码如下: class Solution { public: bool dictContain(unordered_set<string> &dict, string s) { unordered_set<string>::iterator ite = dict.find(s); if(ite != dict.end()) retu…

LeetCode: Word Break II [140]

[题目] Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences. For example, given s = "catsanddog", dict = ["cat", "cat…

[LeetCode] Distinct Subsequences 解题思路

Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative…

LeetCode——Word Break

Question Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, given s ="leetcode", dict =["leet", "code"]. Return true…

[LeetCode]Word Break 特里

意甲冠军:推断字符串给定的字符串是否构成词典. 来推断目标字符串相匹配整个字典.我们需要来推断目标字符串的每个前缀开始的下一场比赛,这需要匹配目标字符串的成功,所有前缀的枚举. class TrieNode{//from http://www.cnblogs.com/x1957/p/3492926.html public: TrieNode* ch[26];//char指针数组 bool isWord; TrieNode():isWord(false){ memset(ch,0,sizeof(T…

[LeetCode] Word Break II (TLE)

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences. For example, givens = "catsanddog",dict = ["cat", "cats"…

[LeetCode] Decode Ways 解题思路

A message containing letters from A-Z is being encoded to numbers using the following mapping: 'A' -> 1 'B' -> 2 ... 'Z' -> 26 Given an encoded message containing digits, determine the total number of ways to decode it. For example,Given encoded…

LeetCode: Word Search 解题报告

Word SearchGiven a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell ma…

[LeetCode] Interleaving String 解题思路

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2. For example,Given:s1 = "aabcc",s2 = "dbbca", When s3 = "aadbbcbcac", return true.When s3 = "aadbbbaccc", return false. 问题 : 给定三个字符串 s1, s…

LeetCode Two Sum 解题思路(python)

问题描述 给定一个整数数组, 返回两个数字的索引, 使两个数字相加为到特定值. 您可以假设每个输入都有一个解决方案, 并且您不能使用相同的元素两次. 方法 1: 蛮力 蛮力方法很简单.循环遍历每个元素 xx 并查找是否有另一个值等于目标 xtarget−x. class Solution: def twoSum(self, nums, target): """ :type nums: List[int] :type target: int :rtype: List[int]…

【LeetCode】140. Word Break II 解题报告（Python & C++）

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 递归求解 日期 题目地址:https://leetcode.com/problems/word-break-ii/ 题目描述 Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to con…

【leetcode】Word Break （middle）

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, givens = "leetcode",dict = ["leet", "code"]. Return true because &…