题解 | #牛群的信息传递#

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param message string字符串 
     * @param keyword string字符串 
     * @return int整型
     */
    int findKeyword(string message, string keyword) 
    {
        // write code here
        vector<vector<int>> dp(keyword.size() + 1,vector<int>(message.size() + 1,0));
        for (int i = 1;i <= keyword.size();++i)
        {
            for (int j = 1;j < message.size();++j)
            {
                if (keyword[i - 1] == message[j - 1])
                {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                }
            }
        }
        int index = -1;
        for (int i = 1;i <= message.size();++i)
        {
            if (dp[keyword.size()][i] == keyword.size())
            {
                index = i;
                break;
            }
        }
        if (index == -1)
        {
            return -1;
        }
        return index - keyword.size();
    }
};

那就借助动态规划吧~~~