leetcode 139. Word break word split (medium)

One 、 The main idea of the topic

label :  Dynamic programming

https://leetcode.cn/problems/word-break

Give you a string  s  And a list of strings  wordDict  As a dictionary . Please judge whether you can use the words in the dictionary to splice  s .

Be careful ： It is not required to use all the words in the dictionary , And the words in the dictionary can be reused .

Example  1：

Input : s = "leetcode", wordDict = ["leet", "code"] Output : true explain :  return  true  because  "leetcode"  Can be  "leet"  and  "code"  Stitching into .

Example  2：

Input : s = "applepenapple", wordDict = ["apple", "pen"] Output : true explain :  return  true  because  "applepenapple"  Can be  "apple" "pen" "apple"  Stitching into .   Be careful , You can reuse the words in the dictionary .

Example  3：

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

Tips ：

• 1 <= s.length <= 300

• 1 <= wordDict.length <= 1000

• 1 <= wordDict[i].length <= 20

• s  and  wordDict[i]  Only lowercase letters

• wordDict  All the strings in   Different from each other

Two 、 Their thinking

The segmentation condition of this problem is determined by the string in the set , Therefore, when considering each division position , You need to traverse the string collection , To determine whether the current position can be successfully segmented , Note for location 0, The initialization value is required to be true .

3、 ... and 、 How to solve the problem

3.1 Java Realization

public class Solution {
 public boolean wordBreak(String s, List<String> wordDict) {
 int n = s.length();
 boolean[] dp = new boolean[n + 1];
 dp[0] = true;
 for (int i = 1; i <= n; i++) {
 for (String tmp : wordDict) {
 int len = tmp.length();
 if (i>=len && tmp.equals(s.substring(i - len, i))) {
 dp[i] = dp[i] || dp[i - len];
 }
 }
 }
 return dp[n];
 }
}


Four 、 Summary notes

• 2022/6/24  It's Friday again , The weekend is fast