[LeetCode] 139. Word Break

题：https://leetcode.com/problems/word-break/description/

题目

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

思路

动态规格
字符串s中，下标i。dp[i] 为位置 i 前的字符都包含在字典中。（i之前，所以在遍历i时需要到len(s)）
状态转移的方程为： dp[i] = True if dp[j] and s[j:i] in wordDict ,j = 0,…,i-1
经过一次 i 遍历便可得到。

note：

1.对于 j 的遍历顺序，如果从0开始，即把 s[0:i] 从前往后切 s[0:j] 、 s[j:i]，看 s[j:i] 是否在 wordDict中，由于整个s存在wordDict中比较少，会增加额外的开销。更好的方法是j从i-1到0遍历，即把s[0:i]从后往前切。

2.python 语法：

#生成 python 数组的方法：
dp = [False]*(len(s))

#生成 python 默认Dcit的方法 （查询的速度会更快）
from collections import defaultdict
def get_False():
    return False
dp = defaultdict(get_False)

code

from collections import defaultdict

def get_False():
    return False

class Solution:
    def wordBreak(self, s, wordDict):
        """
        :type s: str
        :type wordDict: List[str]
        :rtype: bool
        """
        dp = defaultdict(get_False)
        # dp = [False]*(len(s)+1)
        dp[0] = True
        for i in range(1,len(s)+1):
            for j in range(i-1,-1,-1):
                if dp[j] and s[j:i] in wordDict:
                    dp[i] = True
                    break
        return dp[len(s)]