problem:
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.
thinking:
(1)最优解,DP问题
(2)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]));
code:
class Solution { public: int minCut(string s) { int len = s.size(); int* dp = new int[len+1]; for(int i=len; i>=0; i--) dp[i] = len-i; bool** matrix = new bool*[len]; for(int i=0; i<len; i++) { matrix[i] = new bool[len]; memset(matrix[i], false, sizeof(bool)*len); } for(int i=len-1; i>=0; i--) for(int j=i; j<len; j++) { if(s[i] == s[j] && (j-i<2 || matrix[i+1][j-1])) { matrix[i][j] = true; dp[i] = min(dp[i], dp[j+1]+1); } } return dp[0]-1; } };
leetcode || 132、Palindrome Partitioning II
原文地址:http://blog.csdn.net/hustyangju/article/details/45361731