139. Word Break

问题描述
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”.

解题思路
该问题给了我们一个字符串s和一个字典dict，要求我们根据字典判断字符串s能否全部根据字典中的单词进行分割。我们可以这样思考：字符串s可以被全部分割，即组成字符串s的子字符串全部都在字典dict中，而判断子字符串是否在dict中，我们可以一个一个字符的判断，用一个bool类型的数组或者向量，向量或数组的长度为字符串s的长度加1，然后每一位初始设为false，第一位设为true。然后遍历没一个字符，如果该字符在dict中就将对应位置的数组设为true，每一位的判断先要满足前一位为true的条件，然后再进行判断，最后得出的字符串s长度的位置为真即代表可以完全分割。

代码展示

#include<iostream>#include<stdlib.h>#include<stdio.h> #include<string>#include<vector>#include<algorithm>using namespace std;class Solution {public:    bool wordBreak(string s, vector<string>& wordDict) {        int size = s.size();        vector<bool> dp(size + 1, false);        dp[0] = true;        for(int i = 1; i <= size; i++){            for(int j = 0; j < i; j++){                if(dp[j]){                    vector<string>::iterator re = find( wordDict.begin(), wordDict.end(),  s.substr(j, i - j));                     if (re != wordDict.end())                         dp[i] = true;                }            }        }        return dp[size];    }};int main(){    string s;    cout<<"请输入原始字符串s：";    cin>>s;    cout<<"请输入字典向量的额元素个数n：";    vector<string> wordDict;    int n;    string a;    cin>>n;    while(n--){        cin>>a;        wordDict.push_back(a);    }    Solution solution;    bool result=solution.wordBreak(s, wordDict);    if(result == true) cout<<"True"<<endl;    else cout<<"False"<<endl;    return 0;} 

运行结果展示