Palindrome Partitioning II

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题目原型：

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.

基本思路：

题目的意思就是在上一题的基础上找出切分次数最少的组合，即尽可能让回文字符串更长。那么我们同样生成table表用来存储i到j之间的字符串是否是回文；然后遍历字符串来更新最小的切分次数。

public int minCut(String s){if(s==null||s.length()==0)return 0;//创建一个table表boolean[][] table = new boolean[s.length()][s.length()];int[] rs = new int[s.length()];for(int i = 0;i<table.length;i++){for(int j = 0;j<table.length;j++)table[i][j] = true;}//产生table表genTable(table,s);if(table[0][table.length-1])return 0;for(int i = 1;i<s.length();i++){if(table[0][i]){rs[i] = 0;continue;}rs[i] = rs[i-1]+1;//不要s[i]参与//要s[i]参与for(int j = 0;j<i;j++){if(table[j][i])rs[i] = Math.min(rs[i],rs[j-1]+1);}}return rs[rs.length-1];}//产生table表public void genTable(boolean[][] table,String s){//根据间距从小到大来遍历for(int dis = 1;dis<table.length;dis++){for(int i = 0,j = i+dis;j<table.length;i++,j++){table[i][j] = (table[i+1][j-1]&&(s.charAt(i)==s.charAt(j)));}}}

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