題目：

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
s?=?"leetcode",
dict?=?["leet", "code"].

Return true because?"leetcode"?can be segmented as?"leet code".

思路：

1、DFS

2、動態規劃

代碼：

#include<iostream> #include<unordered_set> #include<vector>using namespace std;// dp bool WordBreak_dp(string s,unordered_set<string> &wordDict){int n=s.size();vector<bool> dp(n+1,false);dp[0]=true;for(int i=0;i<n;i++){for(int j=i;j>=0;j--){if(!dp[j]) continue;if(wordDict.find(s.substr(j,i-j+1))!=wordDict.end()){dp[i+1]=true;break;}}}return dp[n];}// dfs bool WordBreak_dfs(string s,unordered_set<string> &wordDict){if(s=="") return true;for(int i=0;i<s.size();i++){if(wordDict.find(s.substr(0,i+1))!=wordDict.end()){if(WordBreak_dfs(s.substr(i+1),wordDict))return true;}}return false; }int main(){unordered_set<string> wordDict;wordDict.insert("leet");wordDict.insert("code");string s;while(cin>>s){cout<< WordBreak_dp(s,wordDict) <<endl;cout<< WordBreak_dfs(s,wordDict) <<endl;}return 0; }

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