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.
class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
set<string> strset(wordDict.begin(), wordDict.end());//vector-->set
if (wordDict.size() == 0)
{
return false;
}
//动态规划
vector<bool> dp(s.size() + 1, false);
dp[0] = true;
for (int i = 1; i <= s.size(); i++) //第一层遍历，s的每个位置是否可分成字典元素
{
for (int j = i - 1; j >= 0; j--)
{
if (dp[j]) //j之前的元素可以分成字典元素
{
if (strset.find(s.substr(j, i - j)) != strset.end())
{
dp[i] = true;
break;
}
}
}
}
return dp[s.size()];
}
};
题目来源
