地址:https://oj.leetcode.com/problems/word-break-ii/ 算法:第二题与第一题不同的是,第二题需要构造出所有可能的解,这样的话,我们就必须把每个子问题的所有解情况都存储起来。用一个二维数组dp来存储每一个子问题的解,其中dp[i]存储了所有满足条件的j。最后利用递归的方法构造出所有的解,看代码应该会比...
public:boolwordBreak(strings,vector<string>& wordDict){if(wordDict.size() ==0|| s.empty())returnfalse;vector<int>dp(s.length() +1,false); dp[0] =true;for(inti =1; i <= s.length(); i++) {for(intj =0; j <= i; j++) {if(dp[j]) {// 第二个参数表示,表示从j开始的...
140. Word Break II Hard Topics Companies Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order. Note that the same word in the dictionary may be reused...
Can you solve this real interview question? Word Break II - Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order. Note
【Leetcode】Word Break II 题目链接:https://leetcode.com/problems/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....
【Leetcode】Word Break 题目链接:https://leetcode.com/problems/word-break/ 题目: 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...
一到源自于 LeetCode 上的题,链接 Word Ladder II - LeetCodeleetcode.com/problems/word-ladder-ii/description/ 根据题意,要找到 beginWord 到 endWord 的所有变换路径,双向搜索的思路是:依次填充以 beginWord 和 endWord 为终点的两个有向图,当两个有向图出现共同单词时,以共同单词为连接点,连接两个有向...
139. Word Breakwindliang 互联网行业 开发工程师 来自专栏 · LeetCode刷题 1 人赞同了该文章 题目描述(中等难度) 给一个字符串,和一些单词,问字符串能不能由这些单词构成。每个单词可以用多次,也可以不用。 解法一 回溯 来一个简单粗暴的方法,利用回溯法,用 wordDict 去生成所有可能的字符串。
dict = ["leet", "code"]. Return true because "leetcode" can be segmented as "leet code". ** My code: importjava.util.Set;publicclassSolution{publicbooleanwordBreak(Strings,Set<String>wordDict){boolean[]isBreakUp=newboolean[s.length()+1];isBreakUp[0]=true;for(inti=1;i<s.length()+...
提交后,leetcode提示超时,看到relate topic提示是动态规划方法。 于是从这个角度想到了这个解法,dp[k]表示从头开始到第k个字符的字符串能否被dict完美break,dp[0]默认为true,可理解为空字符串能够被break。 求解dp[k]是否为true,只需要看0<=i<k这个范围是否存在某处使得dp[i] == true并且s.substring(i, k...