139. Word Break

最新推荐文章于 2025-04-13 18:39:57 发布

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本文探讨了如何通过动态规划解决字符串分割问题，即判断一个非空字符串能否由字典中的一个或多个单词组成。文章提供了详细的算法实现，并附带了一个示例进行说明。

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问题描述
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.
解决思路
使用动态规划，dp[i]表示s[0…i-1]是否能被dict中的词组合表示。dp[0] = true。对于dp[i]，遍历 j (i-1,i-2,…0),若dp[j] = true, 而且s.substr(j,i-j)在dict中，则dp[i] = true。
结果为dp[len]
代码

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        int len = s.length();
        set<string> dict;
        for (auto x : wordDict)
            dict.insert(x);
        bool dp[len+1] = {0};
        dp[0] = true;
        for (int i = 1; i <= len; ++i) {
            for (int j = i-1; j >= 0; --j) {
                if (dp[j]) {
                    string tmp = s.substr(j,i-j);
                    if (dict.find(tmp) != dict.end()) {
                        dp[i] = true;
                    }
                }
            }
        }
        return dp[len];
    }
};