简历：The Google Resume: How to Prepare for a Career and Land a Job at Apple, Microsoft, Google, or any Top Tech Company
算法学习书籍：Introduction to Algorithms
编程珠玑：Programming Pearls (2nd Edition)
C++ 学习：The C++ Programming Language, 4th Edition
经典操作系统书籍，龙书：Operating System Concepts
创业：The Start-up of You: Adapt to the Future, Invest in Yourself, and Transform Your Career
Given a string s, partition s such that every substring of the partition is a palindrome.
Return the minimum cuts needed for a palindrome partitioning of s.
For example, given s =
Return
» Solve this problem"aab",Return
1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.
刚开始还是用的回溯，和I一样，这样可以做，但是会超时在跑大数据的时候。然后参考了“水中的鱼”的博客，用的DP。并且他总结，一般求最优的时候，都是用dp，不过要自己设计dp函数和情况，这道题靠裸想还是有些难度的，如果对dp不熟悉的话。
首先要定义一个函数D[i]=min(D[i], D[j+1]+1). 假如此时 P[i][j]是一个palindrome。记得要初始化，相当于就有两个指针，i, j，i从s.size()处开始，j从i开始往s.size()处数，这样最后D[0]-1的值就是最小的cut 数。
所以这个时候我们还要定义一个P[i][j]=true if S[i]==S[j]&&(j-i<2 || P[i+1][j-1])
记得初始化！
No comments:
Post a Comment