leetcode -- Palindrome Partitioning II

最新推荐文章于 2024-05-28 06:45:00 发布

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最新推荐文章于 2024-05-28 06:45:00 发布

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原文链接：http://www.cnblogs.com/taizy/p/4008836.html

本文介绍了一种算法，用于计算将字符串分割成多个回文子串所需的最少切割次数。通过动态规划的方法，预先计算字符串各部分是否为回文，并在此基础上递推求解最小分割数。

指责别人，看清自己

[问题描述]

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 = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.

[解题思路]

ans[i]表示s[i]至最后需要的划分数

注意：line6 "<="

 1 int Solution::minCut(string s){
 2     const int len = s.length();
 3     bool pali[len][len];
 4     int ans[len];
 5     fill_n(&pali[0][0], len*len, false);
 6     for (int i = 0; i <= len; i++)
 7         ans[i] = len - i - 1;
 8     for (int i = len - 1; i >= 0; i --){
 9         for (int j = i; j < len; j ++){
10             pali[i][j] = s[i]==s[j]&&(j-i<2||pali[i+1][j-1]);
11             pali[i][j] == true?(ans[i] = min(ans[i], ans[j + 1] + 1)):(true);
12         }
13     }
14     return ans[0];
15 }