LeetCode——Word Break

2024-04-03 21:40:05

LeetCode——Word Break

Question

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".

Solution

动态规划的思想，从头开始扫描字符串，判断当前子字符串是否可以在字典中查找到，取决于自身以及对这个子字符串的所有划分方式是否可以查找到。
递推关系式子： dp[i] = dp[j] && dp[i - j] (0 <= j <= i)
初始值 dp[0] = true; 表示子字符串长度为0的时候，是可以查找到的。

Answer

public:bool wordBreak(string s, vector<string>& wordDict) {if (wordDict.size() == 0 || s.empty())return false;vector<int> dp(s.length() + 1, false);dp[0] = true;for (int i = 1; i <= s.length(); i++) {for (int j = 0; j <= i; j++) {if (dp[j]) {// 第二个参数表示，表示从j开始的字符个数string str1 = s.substr((unsigned int)j, i - j);if (check(wordDict, str1)) {dp[i] = true;break;}}}}return dp[s.length()];}bool check(vector<string>& wordDict, string& str1) {for (string str : wordDict)if (str == str1)return true;return false;}
};

转载于:https://www.cnblogs.com/zhonghuasong/p/6957705.html

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