[leetcode] Word Break

最新推荐文章于 2023-06-01 11:39:37 发布

Cabinathor

最新推荐文章于 2023-06-01 11:39:37 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/lmxmimihuhu/article/details/46490195

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本文详细阐述了解决LeetCode中Word Break问题的算法实现，包括动态规划方法的应用及代码解析。

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From : https://leetcode.com/problems/word-break/

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

class Solution {
public:
    bool wordBreak(string s, unordered_set<string> &dict) {
        int n = s.size();
        // dp[i]：i之前（不含i）已保证匹配成功
		vector<bool> dp(n+1, false);
		dp[0]=true;
		for(int i=0;i<n;i++) {
		    // 查看0~i是否可以匹配
			if(dp[i]) {
				int idx=i;
				for(int j=idx;j<n;j++) {
					string cur=s.substr(idx,j-idx+1);
					if(dict.find(cur) != dict.end())
						dp[j+1]=true;
				}
			}
		}
		return dp[n];
	}
};

public class Solution {
    public boolean wordBreak(String s, Set<String> wordDict) {
        if (s == null || wordDict == null) {
			return false;
		}
		int n = s.length();
		// dp[i]：i之前（不含i）已保证匹配成功
		boolean[] containTo = new boolean[n + 1];
		containTo[0] = true;
		for (int i = 0; i < n; ++i) {
			// 查看0~i是否可以匹配
			if (containTo[i]) {
				for (int j = i; j < n; j++) {
					String cur = s.substring(i, j + 1);
					if (wordDict.contains(cur)) {
						containTo[j + 1] = true;
					}
				}
			}
			if (containTo[n] == true) {
				return true;
			}
		}
		return containTo[n];
    }
}