【leetcode每日刷题】【DP】139. Word Break

本文链接： https://blog.csdn.net/m0_38103546/article/details/100812148

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.

Note:

The same word in the dictionary may be reused multiple times in the segmentation.
You may assume the dictionary does not contain duplicate words.

Example 1:

Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
             Note that you are allowed to reuse a dictionary word.

Example 3:

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

import java.util.ArrayList;
import java.util.Arrays;

class num139 {
    public boolean wordBreak(String s, ArrayList<String> wordDict) {
        int len = s.length();
        boolean[] dp = new boolean[len+1];
        dp[0] = true;
        for(int i=1; i<=len; i++){
            for(int j=i-1; j>=0; j--){
                if(dp[j] && wordDict.contains(s.substring(j, i))){
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[len];
    }
    public static void main(String[] args) {
        num139 solution = new num139();
        String s = "leetcode";
        ArrayList<String> wordDict = new ArrayList<String>(Arrays.asList("leet", "code"));
        boolean result = solution.wordBreak(s, wordDict);
        System.out.println(result);
    }
}

使用动态规划的方法，使用数组记忆从起始元素开始到当前位置是否可以在字典中找到，分为两种情况，一种是从index 0 开始到当前位置的子串在字典中，另一种情况是从已经在字典中的子串+1的位置开始到当前位置的子串在字典中，如果遍历到最后，第len+1的位置上记忆数组值为true,则返回true.

 s = "leetcode", wordDict = ["leet", "code"]

l	e	e	t	c	o	d	e
0	0	0	0	1	0	0	0	1

【leetcode每日刷题】【DP】139. Word Break

猜你喜欢