leetcode139 Word Break

"""
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
"""
"""
动态规划：类似于322题：https://www.cnblogs.com/yawenw/p/12298704.html
方程为dp[i] = { True if dp[:i] in dict 或者 dp[j] if dp[j:i] in dict }
传送门：https://blog.csdn.net/daniel_hh/article/details/89575491
"""
class Solution:
    def wordBreak(self, s, wordDict):
        dp = [False] * (len(s) + 1)
        dp[0] = True
        if s in wordDict:
            return True
        for i in range(1, len(s) + 1):  # 从第一个到最后一个字符
            for j in range(i):  # i之前的第一个到i个字符
                #切片的用法
                #nums = [2, 5, 8, 4]  print(nums[0:3]) == [2, 5, 8]
                #string = "abcdefg"   print(string[5:110]) == 'fg'
                if dp[j] and s[j:i] in wordDict:  # !!!动态方程
                    dp[i] = True
                    break #break 可有可无
        return dp[len(s)]