解题思路:这题用到两种方法,不过其实是一个很巧妙的二重dp 1)暴力求解以i开头,j结尾的是否为回文,如果是,ans += sum[i-1](ans 为答案, sum[i]为在 1 - i内回文串的个数--需要dp求解) 这里非常耗时,时间大约为 n^3 , 跑出来为 830ms 解题代码: View Code 2) 这里在前面一种解法上又多了一次dp...
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