[leetcode]139. Word Break单词能否拆分

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".

题意：

给定字符串S和字典wordDict, 判断字符串S是否能由字典中的单词组合而成

思路：

用一维boolean数组记录是否能字符串S是否能被分隔

l       e      e      t       c      o      d     e

T	F	F	F	F	F	F	F	F
0	1	2	3	4	5	6	7	8

dp[0]初始化为true, 其他为default的false

dp[s.length() + 1]

i = 1         j = 1,  dp[1] 为default F, 没必要再去验证wordDict.contains(s.substring(j,i))

j = 0,  dp[0] 为 T ， 验证wordDict.contains(s.substring(0,1)) 为F

i = 2        j = 2,  dp[2] 为default F

j = 1,  dp[1] 为default F

　　　　  j = 0,  dp[0] 为 T ， 验证wordDict.contains(s.substring(0,2)) 为F

i = 3        j = 3,  dp[3] 为default F

j = 2,  dp[2] 为default F

j = 1,  dp[1] 为default F

　　　　  j = 0,  dp[0] 为 T ， 验证wordDict.contains(s.substring(0,3)) 为F

i = 4       j = 4, dp[4]为default F

　　　　 j = 3,  dp[3] 为default F

j = 2,  dp[2] 为default F

j = 1,  dp[1] 为default F

　　　　  j = 0,  dp[0] 为 T ， 验证wordDict.contains(s.substring(0,4)) 为T  ----> 更新dp[4] = true

T	F	F	F	T	F	F	F	F
0	1	2	3	4	5	6	7	8

代码：

 class Solution {
     public boolean wordBreak(String s, List<String> wordDict) {
         // corner case
         if (s == null || s.length() < 1 || wordDict == null || wordDict.size() < 1) {
             return false;
         }
         boolean[] dp = new boolean[s.length() + 1];
         dp[0] = true;
         //外层循环scan字符串S
         for(int i = 1; i <= s.length(); i++ ){
             //内层循环找分割点
             for(int j = i; j >= 0; j--){
                 if(dp[j] && wordDict.contains(s.substring(j,i))){
                     dp[i] = true;
                 }
             }
         }
         return dp[s.length()];
     }
 }