Word Break

这个题目思路：在一个bool型数组中，像接力一样传递匹配成功，传递到最后一个字符，说明匹配成功。说的明白点就是从第i（0~n）个字符开始向后与子串进行匹配，匹配的数组中标记为true，循环比较。

需要注意的是：unordered_set的count（T s）查看是否包含该元素。string类的substr（pos，n）函数返回从pos位置开始的n个字符。

 1 class Solution {
 2 public:
 3     bool wordBreak(string s, unordered_set<string> &dict) {
 4         int MainLen = s.size();
 5         vector<bool> dp(MainLen,false);
 6         dp[0] = true;
 7         for(int i=0;i<MainLen;i++)
 8         {
 9             if(dp[i])
10             {
11                 for(int j = i;j<MainLen;j++)
12                 {
13                     if(dict.count(s.substr(i,j-i+1)))
14                     {
15                         dp[j+1] = true;
16                     }
17                 }
18             }
19         }
20         return dp[MainLen];
21     }
22 };

时间： 2024-10-04 21:37:05

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