LeetCode 139. Word Break

139. Word Break

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• Total Accepted: 131482
• Total Submissions: 457810
• Difficulty: Medium
• Contributors: Admin

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".

UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

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【题目分析】

给定一个目标字符串和一个字符串列表，判断目标字符串是否能由字符串列表中的串组合而成。

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【思路】

1. DP套路多，把大问题分解为子问题还是很考验难度的。

2. 要认真分析，不要急躁。

3. 要注意下标

4. 在本问题中，判断一个大的字符串是否可以被组合成功，可以分解为s.substring(0,i)和substring(i, s.length()),(0=<i<s.length())是否存在一个i使得被分成两部分的字符串都可以被列表中的串组合成功。

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【java代码】

public class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        boolean[] flag = new boolean[s.length()+1];
        flag[0] = true;
        
        for(int i = 1; i <= s.length(); i++) {
            for(int j = i-1; j >= 0; j--) {
                if(flag[j] && wordDict.contains(s.substring(j,i))) {
                    flag[i] = true;
                    break;
                }
            }
        }
        
        return flag[s.length()];
    }
}