Leetocde: Palindrome Partitioning II

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我是靠谱客的博主疯狂服饰，最近开发中收集的这篇文章主要介绍Leetocde: Palindrome Partitioning II，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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.

方法一DFS: Time Limit Exceeded

void dfs(string s, int depth, int &Min)
{
if(s.size() < 1)
{
if(Min > depth-1) Min = depth-1;
return;
}
for(int i = s.size()-1; i >= 0; i--)
{
if(s[i] != s[0])continue;
int begin = 0;
int end = i;
while(begin < end)
{
if(s[begin] == s[end])
{
begin++;
end--;
}
else
break;
}
if(begin >= end)
dfs(s.substr(i+1), depth+1, Min);
}
}
int minCut(string s) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
int Min = s.size();
vector<string> path;
dfs(s,0,Min);
return Min;
}

方法二DP: Accepted

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]));

int minCut(string s) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
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;
}