leetCoder-wordBreak判断能否分词

题目

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words.

For example, given
s = “leetcode”,
dict = [“leet”, “code”].
Return true because “leetcode” can be segmented as “leet code”.
UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

分析

给出一段话,判断能否分词(被dict里的单词分割)

简单动态规划题,使用dp[i]表示前i个单词能否被分词,则状态转移方程为

dp[j] = dp[i] && s[i:j]∈dict

如果前i个单词可以被分词,且i-j在dict里,则前j个单词可以被分词

AC代码

class Solution {public:    bool wordBreak(string s, vector<string>& wordDict) {        int n = s.length();        vector<bool> dp(n + 1, false);        dp[0] = true;        for(int i=0;i<n;i++){            for(int j = i; dp[i]&&j<n; j++){                auto f = find(wordDict.begin(), wordDict.end(), s.substr(i,j-i+1));                if(f != wordDict.end())//找到                    dp[j+1] = true;            }        }        return dp[n];    }};