leetcode Distinct Subsequences

最新推荐文章于 2017-07-04 10:43:23 发布

原创最新推荐文章于 2017-07-04 10:43:23 发布 · 377 阅读

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本文采用动态规划方法解决字符串匹配问题，详细介绍了算法实现过程，并通过实例代码展示了解决方案。

此题利用动态规划

参考博客http://blog.youkuaiyun.com/abcbc/article/details/8978146

代码

class Solution {
public:
    int numDistinct(string S, string T) {
        int sLen = S.length();
        int tLen = T.length();
        int **numArray;
        numArray = new int*[tLen+1];
        for(int i = 0; i <=tLen; ++i)
            numArray[i] = new int[sLen+1];
        
        numArray[0][0] = 1;
        for(int i = 1; i <=tLen; ++i)
            numArray[i][0] = 0;
        for(int i = 1; i <= sLen; ++i)
            numArray[0][i] = 1;
            
        for(int i = 1; i <= tLen; ++i)
           for(int j = 1; j <= sLen; ++j)
           {
               numArray[i][j] = numArray[i][j-1];
               if(T[i-1]==S[j-1])
                  numArray[i][j] += numArray[i-1][j-1];
               
           } 
           
           return numArray[tLen][sLen];
    }
};