139. Word Break

问题：

给定字符串s，和词典wordDict

若s可拆分成多个子串，存在于词典中，则返回true。否则返回false。

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

Constraints:
1 <= s.length <= 300
1 <= wordDict.length <= 1000
1 <= wordDict[i].length <= 20
s and wordDict[i] consist of only lowercase English letters.
All the strings of wordDict are unique.

　　

解法：DP

1.状态：

• i： 字符串s[0~i]

2.选择：

• 字符串s[0~i]是否在词典中？
• 字符串s[1~i]是否在词典中 | | s[0]是否在词典中 ？
• 字符串s[2~i]是否在词典中 | | s[0~1]是否在词典中 ？
• ......
• 字符串s[i]是否在词典中 | | s[0~i-1]是否在词典中 ？

以上任意满足，则s[0~i]是符合题意的字符串。

3.状态转移：

• dp[i+1]：字符串s[0~i]是满足题意的字符串，即可拆分为词典单词。
• = OR {   j=0~i
  • dp[j] && s[j~i] in wordDict
  • }

4.base：

dp[0]=true

 

代码参考：

class Solution {
public:
    //dp[i+1]:s[0~i]: can be matched ?
    //opt: OR :
    //case_1: dp[0] && s[0~i] in wordDict
    //case_2: dp[1] && s[1~i] in wordDict
    //...
    //case_n: dp[i] && s[i] in wordDict
    //base: dp[0]=true
    bool wordBreak(string s, vector<string>& wordDict) {
        int n = s.length();
        unordered_set<string> wD(wordDict.begin(), wordDict.end());
        vector<bool> dp(n+1, false);
        dp[0]=true;
        for(int i=1; i<=n; i++) {
            for(int j=0; j<=i; j++) {
                if(dp[j]) {
                    string sub = s.substr(j, i-j);
                    dp[i] = (wD.count(sub)>0);
                    if(dp[i]) break;
                }
            }
        }
        return dp[n];
    }
};