Given a string s and a dictionary of words dict, determine if s can be break into a space-separated sequence of one or more dictionary words.

Example

Given s = "lintcode", dict = ["lint", "code"].

Return true because "lintcode" can be break as "lint code".

Solution:

DP.

bool wordBreak(string s, unordered_set<string> &dict) {
int len = s.size();
if(len == ) return true; int min_len = INT_MAX, max_len = INT_MIN;
for (auto &ss : dict) {
min_len = min(min_len, (int)ss.length());
max_len = max(max_len, (int)ss.length());
} vector<int> breakIdx;
breakIdx.push_back(-); for(int i = ;i < len;++i){
for(int j = breakIdx.size() - ;j >= ;--j){
int idx = breakIdx[j];
if(i - idx < min_len || i - idx > max_len)
continue;
string str = s.substr(idx + , i - idx);
if(dict.find(str) != dict.end()){
breakIdx.push_back(i);
break;
}
}
}
return (*breakIdx.rbegin()) == (len - );
}

[LintCode] Word Break的更多相关文章

  1. [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 ...

  2. 【leetcode】Word Break II

    Word Break II Given a string s and a dictionary of words dict, add spaces in s to construct a senten ...

  3. 17. Word Break && Word Break II

    Word Break Given a string s and a dictionary of words dict, determine if s can be segmented into a s ...

  4. 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 ...

  5. 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  ...

  6. 【LeetCode OJ】Word Break II

    Problem link: http://oj.leetcode.com/problems/word-break-ii/ This problem is some extension of the w ...

  7. Leetcode#139 Word Break

    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 ...

  8. 【leetcode】Word Break (middle)

    Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separa ...

  9. 140. Word Break II

    题目: Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where e ...

随机推荐

  1. sqlserver2012相关资源下载

    1.输入网址http://www.codeplex.com 2.找到Microsoft SqlServer Product Samples选项 3.进入之后显示如下页面 4.选择SQL Server ...

  2. cpu和内存的关系

    CPU是负责运算和处理的,内存是交换数据的.当程序或者操作者对CPU发出指令,这些指令和数据暂存在内存里,在CPU空闲时传送给CPU,CPU处理后把结果输出到输出设备上,输出设备就是显示器,打印机等. ...

  3. HNU 12827 NASSA’s Robot

    题目链接:http://acm.hnu.cn/online/?action=problem&type=show&id=12827&courseid=268 #include&l ...

  4. grep与egrep

    当只有一个匹配条件时:egrep pattern file等价于grep -E pattern file 例如: 当多个匹配条件时,只能用egrep -e pattern1 -e pattern2 - ...

  5. mysql中binary相加的问题

    我在mysql中有这样一段代码 SQL code   ? 1 2 3 4 5 6 7 8 declare @byte1 binary(1) declare @byte2 binary(1) decla ...

  6. 【转】基于LDA的Topic Model变形

    转载自wentingtu 基于LDA的Topic Model变形最近几年来,随着LDA的产生和发展,涌现出了一批搞Topic Model的牛人.我主要关注了下面这位大牛和他的学生:David M. B ...

  7. MyBatis3: There is no getter for property named 'code' in 'class java.lang.String'

    mybatis3  : mysql文如下,传入参数为string类型时‘preCode’,运行报错为:There is no getter for property named 'preCode' i ...

  8. 【JAVA、C++】LeetCode 003 Longest Substring Without Repeating Characters

    Given a string, find the length of the longest substring without repeating characters. For example, ...

  9. 【2016-08-06】QTableWidget的一些用法总结

    1. QTableWidget的列宽如何自适应显示区域大小? QTableWidget的列头继承自QHeaderView,因此如果不使用固定列宽而需要Table中多列的列宽自适应显示区域大小的话, 可 ...

  10. Oracle Undo与脏读解析

    Undo就是用来记录保存事务操作过程中的数据,如果事务发生错误,可以之前的数据进行填补. Undo segment 是保存在表空间上的.Undo 大小是固定的,既然是固定的也就是有限的.如果保存的记录 ...