LeetCode 139. Word Break

139. Word Break

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.

/*
题意：给定字符串s, 问是否能够由字典dict中的单词组成
思路：dp[j]表示前j个字符可以由字典dict中的单词组成,
对于长度为i的字符串，若dp[j]为真且子串s.substr(j, i-j)存在于字典dict中，
则长度为i的字符串能够由字典dict中的单词组成（j >= 0 && j < i）
*/
class Solution {
public:
bool find(vector<string>& wordDict, string s){
//用来判断字符串s是否在字典dict中
for(int k = 0; k < wordDict.size(); k++){
if(s == wordDict[k])
return true;
}
return false;
}
bool wordBreak(string s, vector<string>& wordDict) {
vector<bool>dp(s.length() + 1, false);
dp[0] = true;
for(int i = 1; i < s.length() + 1; i++){
for(int j = i - 1; j >= 0; j--){
if(dp[j] && find(wordDict, s.substr(j, i-j))){
//只要找到一种切分方式就说明长度为i的字符串可以成功切分
dp[i] = true;
break;
}
}
}
return dp[s.length()];
}
};