原题地址:https://leetcode-cn.com/problems/word-break-ii/description/
题目描述:
给定一个非空字符串 s 和一个包含非空单词列表的字典 wordDict,在字符串中增加空格来构建一个句子,使得句子中所有的单词都在词典中。返回所有这些可能的句子。
说明:
分隔时可以重复使用字典中的单词。
你可以假设字典中没有重复的单词。
示例 1:
输入:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
输出:
[
"cats and dog",
"cat sand dog"
]
示例 2:
输入:
s = "pineapplepenapple"
wordDict = ["apple", "pen", "applepen", "pine", "pineapple"]
输出:
[
"pine apple pen apple",
"pineapple pen apple",
"pine applepen apple"
]
解释: 注意你可以重复使用字典中的单词。
示例 3:
输入:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
输出:
[]
解题方案:
自己太菜了,不会这题。这道题要用到动态规划和回溯算法,置顶了以便以后学习。
class Solution {
public:
vector<bool> dp;
vector<string> ans, tmp;
void dfs(string &s, vector<string> wordDict, int index) {
if (index == -1) {
string str;
auto it = tmp.rbegin();
str += *it;
++it;
for (; it != tmp.rend(); ++it)
str += " " + *it;
ans.push_back(str);
return;
}
for (auto &x : wordDict)
if (1 + index >= x.size() && (1 + index == x.size() ||
dp[index - x.size()]) && x == string(s, index + 1 - x.size(), x.size())){
tmp.push_back(x);
dfs(s, wordDict, index - x.size());
tmp.pop_back();
}
}
vector<string> wordBreak(string s, vector<string>& wordDict) {
dp.assign(s.size(), false);
for (int i = 0; i < s.size(); ++i) {
for (auto &x : wordDict)
if (x.size() <= i + 1 && (i + 1 == x.size() ||
dp[i - x.size()]) && x == string(s, i + 1 - x.size(), x.size())){
dp[i] = true;
break;
}
}
dfs(s, wordDict, s.size() - 1);
return ans;
}
};