leetcode:Word Break
2014-09-11 16:48
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题目:https://oj.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, given
s =
dict =
Return true because
题目意思是:给定一个字符串和一个字典,判断该字符串能否由一个或者多个字典中的单词组成。
可以用DP的方法解决。
用数组wb[i]保存从0开始到长度为i的子串能否被分割,所以最后的结果返回wb[s.size() - 1]。
首先wb[0] = true.则wb[1]可分割的条件是:wb[0] == true且子串s[0]包含在字典中...wb[i] == true的条件是:存在0 <= j < i,使得从开始长度为j的子串可分割且从j到i的子串存在在字典中...以此类推求出所有的wb[i]...
AC代码:
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"can be segmented as
"leet code".
题目意思是:给定一个字符串和一个字典,判断该字符串能否由一个或者多个字典中的单词组成。
可以用DP的方法解决。
用数组wb[i]保存从0开始到长度为i的子串能否被分割,所以最后的结果返回wb[s.size() - 1]。
首先wb[0] = true.则wb[1]可分割的条件是:wb[0] == true且子串s[0]包含在字典中...wb[i] == true的条件是:存在0 <= j < i,使得从开始长度为j的子串可分割且从j到i的子串存在在字典中...以此类推求出所有的wb[i]...
AC代码:
class Solution { public: bool wordBreak(string s, unordered_set<string> &dict) { int len = s.size(); //创建数组wb[i]表示从0开始长度为i的子串能否被分割;最后的答案则是wb[len] vector<bool> wb(len + 1,false); wb[0] = true; for(int i = 1;i <= s.size();i++){ for(int j = i - 1;j >= 0;j--){ //如果从开始长度为j的子串可以分割而且从j到i的子串出现在字典中 //则从开始长度为i的子串也可以分割 if(wb[j] && dict.count(s.substr(j,i - j))){ wb[i] = true; break; } } } return wb[len]; } };
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