1.代码

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        int length = s.size();
        bool *count1 = new bool[length+1];
        fill(count1, count1 + length + 1, false);
        unordered_map<string, bool> hashTable;
        count1[0] = true;
        for(int i = 0;i < wordDict.size();i ++){
            hashTable[wordDict[i]] = true;
        }
        for(int i = 1;i <= length;i ++){
            for(int j = 0;j < i;j ++){
                if(count1[j] && hashTable.count(s.substr(j,i-j))>0){
                    count1[i] = true;
                    break;
                }
            }
        }
        return count1[length];
    }
};

2.思路

用了动态规划,用一个count1数组记录字符串的前i个字母是否能拆分成单词,状态转移方程是count1[i] = count1[j] &&hashTable.count(s.substr(j,i-j))>0。

08-29 18:12