【LeetCode】140. 单词拆分 II结题报告 (C++)

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 dp;
    vector ans, tmp;

    void dfs(string &s, vector 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 wordBreak(string s, vector& 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;
    }
};


 

你可能感兴趣的:(【LeetCode】140. 单词拆分 II结题报告 (C++))