默认情况下,如果同一行中某个单词太长了,它就会被默认移动到下一行去: word break(normal | break-all | keep-all):表示断词的方式 word wrap(normal | break-word):表示是否要断词 word wrap break-word   [要断词] 独占一行(默认情况下单词太长就会被换到下一行去,所以就独占一行了)的单词被断开成多行, 默认值normal,则不断词,而是一行显示,超出容器 word break break-all:和上面相比…

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…

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",…

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, givens = "catsanddog",dict = ["cat", &q…

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, given s = "leetcode", dict = ["leet", "code"]. Return t…

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<…

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…

Problem link: http://oj.leetcode.com/problems/word-break-ii/ This problem is some extension of the word break problem, so the solution is based on the discussion in Word Break. We also use DP to solve the problem. In this solution, A[i] is not a bool…

原题地址 与Word Break II(参见这篇文章)相比,只需要判断是否可行,不需要构造解,简单一些. 依然是动态规划. 代码: bool wordBreak(string s, unordered_set<string> &dict) { ; for (auto w : dict) maxLen = maxLen > w.length() ? maxLen : w.length(); vector<, false); res[s.length()] = true; ;…

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 &…