Leetcode-139. Word Break

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.

Subscribe to see which companies asked this questio

AC代码1：

class Solution {public:bool wordBreak(string s, vector<string>& wordDict) {int size = wordDict.size();set<string>ma;for (vector<string> ::iterator iter = wordDict.begin(); iter != wordDict.end(); iter++)ma.insert((*iter));int len = s.size();vector<bool>re(len, false);string temp;for (int i = 0; i < len; i++){for (int j = i; j >= 0; j--){if (re[j] == true){ temp = s.substr(j + 1, i - j);if (ma.find(temp) != ma.end()){re[i] = true;break;}}if (j == 0){temp = s.substr(0, i + 1);if (ma.find(temp) != ma.end()){re[i] = true;break;}}}}return re[len - 1];}};

AC代码2：

 class Solution { public: bool wordBreak(string s, vector<string>& wordDict) { int size = wordDict.size(); set<string>ma; for (vector<string> ::iterator iter= wordDict.begin();iter!=wordDict.end(); iter++)ma.insert((*iter)); int len = s.size(); vector<bool>re(len+1,false); re[0] = true; string temp; for (int i = 1; i <=len;i++){ for (int j = i-1; j >= 0;j--){ if (re[j] == true){ temp = s.substr(j, i - j); if (ma.find(temp) != ma.end()){ re[i] = true; break; } } } } return re[len]; } };

解析：

AC代码1是自己写的第一个版本，这时需要注意的的一种情况就是当前下标为i的字母要包含下标为0的字母所组成的字符串存在于wordDict中；

AC代码2利用一个额外的存储空间并将re[0]设置为true，这样的好处就是将AC代码1中的特殊情况统一处理，也就是当j递减到0时，那么这时字符串s就会将第

一个字母包含就去，这种统一化处理方式比较妙！