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.
[Thoughts]
凡是求最优解的,一般都是走DP的路线。这一题也不例外。首先求dp函数,
定义函数
D[i,n]
= 区间[i,n]之间最小的cut数,n为字符串长度
a b a
b b b a b b a b
a
i
n
如果现在求[i,n]之间的最优解?应该是多少?简单看一看,至少有下面一个解
a
b a b b b a b b a
b a
i
j j+1
n
此时 D[i,n] = min(D[i, j] + D[j+1,n])
i<=j
<n。这是个二维的函数,实际写代码时维护比较麻烦。所以要转换成一维DP。如果每次,从i往右扫描,每找到一个回文就算一次DP的话,就可以转换为
D[i]
= 区间[i,n]之间最小的cut数,n为字符串长度, 则,
D[i] = min(1+D[j+1] )
i<=j
<n
有个转移函数之后,一个问题出现了,就是如何判断[i,j]是否是回文?每次都从i到j比较一遍?太浪费了,这里也是一个DP问题。
定义函数
P[i][j]
= true if [i,j]为回文
那么
P[i][j] = str[i] ==
str[j] && P[i+1][j-1];
基于以上分析,实现如下:
public class Solution { public int minCut(String s) { int len = s.length(); int[] minCut = new int[len+1]; boolean[][] isPalindrome = new boolean[len][len]; for(int i = 0; i <= len; i++){ minCut[i] = len -i; } for(int i = 0; i< len; i++){ for(int j = 0; j < len; j++){ isPalindrome[i][j] = false; } } for(int i = len -1; i >= 0; i--){ for(int j =i; j<len; j++){ if(s.charAt(i) == s.charAt(j) && (j-i<2 || isPalindrome[i+1][j-1])){ isPalindrome[i][j] =true; minCut[i] = Math.min(minCut[i], minCut[j+1]+1); } } } return minCut[0]-1; } }
Ref: http://fisherlei.blogspot.com/2013/03/leetcode-palindrome-partitioning-ii.html