2
#yyds干货盘点# LeetCode 热题 HOT 100：单词拆分
source link: https://blog.51cto.com/u_13321676/5780409
Go to the source link to view the article. You can view the picture content, updated content and better typesetting reading experience. If the link is broken, please click the button below to view the snapshot at that time.
#yyds干货盘点# LeetCode 热题 HOT 100：单词拆分
精选 原创给你一个字符串 s 和一个字符串列表 wordDict 作为字典。请你判断是否可以利用字典中出现的单词拼接出 s 。
注意：不要求字典中出现的单词全部都使用，并且字典中的单词可以重复使用。
输入: s = "leetcode", wordDict = ["leet", "code"]
输出: true
解释: 返回 true 因为 "leetcode" 可以由 "leet" 和 "code" 拼接成。
输入: s = "applepenapple", wordDict = ["apple", "pen"]
输出: true
解释: 返回 true 因为 "applepenapple" 可以由 "apple" "pen" "apple" 拼接成。
注意，你可以重复使用字典中的单词。
输入: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
输出: false
代码实现：
public class Solution {
public boolean wordBreak(String s, List<String> wordDict) {
Set<String> wordDictSet = new HashSet(wordDict);
boolean[] dp = new boolean[s.length() + 1];
dp[0] = true;
for (int i = 1; i <= s.length(); i++) {
for (int j = 0; j < i; j++) {
if (dp[j] && wordDictSet.contains(s.substring(j, i))) {
dp[i] = true;
break;
}
}
}
return dp[s.length()];
}
}
public boolean wordBreak(String s, List<String> wordDict) {
Set<String> wordDictSet = new HashSet(wordDict);
boolean[] dp = new boolean[s.length() + 1];
dp[0] = true;
for (int i = 1; i <= s.length(); i++) {
for (int j = 0; j < i; j++) {
if (dp[j] && wordDictSet.contains(s.substring(j, i))) {
dp[i] = true;
break;
}
}
}
return dp[s.length()];
}
}
Recommend
About Joyk
Aggregate valuable and interesting links.
Joyk means Joy of geeK