leetcode 139 Word Break

最新推荐文章于 2025-04-13 18:39:57 发布

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本文探讨了一种使用动态规划解决字符串分割问题的方法，通过检查一个字符串是否能由字典中的单词构成，允许单词重复使用。文章详细解释了算法流程，并提供了代码实现。

描述：一个非空字符串s，包含一系列非空单词的字典序列，判断s是否可由一个或多个电词表中的元素构成

Note:

The same word in the dictionary may be reused multiple times in the segmentation.
You may assume the dictionary does not contain duplicate words

Example 1:

Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code"

Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple"
Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false

这是一道可重复取用的动态规划问题：

为了保证有一个初始的true值，dp[0] = true; dp的第i位对应字符串s的第i-1位；

class Solution {
public:
    static bool cmp(string a, string b)
    {
        return a.size() < b.size();
    }
    bool wordBreak(string s, vector<string>& wordDict) {
       
        sort(wordDict.begin(), wordDict.end(), cmp);
        int len = s.size();
        int size = wordDict.size();
        if(size == 0)
            return false;
        vector<int>dp(len + 1, 0);
        dp[0] = 1;
        for(int i = 1; i <= len; i++)
        {
            for(int j = 0; j < size; j++)
            {
                int sizestr = wordDict[j].size();
                if(i >= sizestr)
                {
                    string str = s.substr(i - sizestr, sizestr);
                    if(str == wordDict[j])
                        dp[i] = dp[i - sizestr] || dp[i];
                }
            }
        }
        return dp[len];
    }
};