【LeetCode】Palindrome Partitioning II

题目描述：

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.

class Solution {
public:
vector<vector<int>> pal;
vector<vector<string>> res;
int min;
int minCut(string s)
{
int path = 0;
min = s.length();
for (int i = 0; i < s.size(); i++)
{
vector<int> line;
pal.push_back(line);
}
for (int i = 0; i < s.length(); i++)
ispal(s, 0, i);
for (int i = 1; i < s.length(); i++)
{
ispal(s, i, s.length() - 1);
}
cut(0, path, s);
return min;
}
void cut(int start, int path, string &s)
{
if (start>s.length() - 1)
{
min = (path - 1)<min ? (path - 1) : min;
return;
}
for (int i = 0; i < pal[start].size(); i++)
{
int temp = path;
temp++;
cut(pal[start][i], temp, s);
}
}
bool ispal(string &s, int start, int end)
{
if (start > end)
return true;
if (ispal(s, start + 1, end - 1) && s[start] == s[end])
{
pal[start].push_back(end + 1);
return true;
}
else
return false;
}
};

但是这段程序会TLE，作为一个初学者实在想不到更好的办法了(￣_￣|||) ，于是去翻了下讨论区，发现新的解法：

class Solution {
public:
int minCut(string s) {
int N = s.size();
vector<int> dp(N+1,numeric_limits<int>::max());
dp[0] = 0;
for(int i=0;i<N;i++)
for(int p=0;p<=1;p++)
for(int l=0;l<=N;l++) {
int a=i-l, b = i+l+p;
if( a < 0 || b >= N || s[a]!=s[b] )
break;
dp[b+1] = min(dp[b+1],dp[a]+1);
}
return dp
-1;
}
};

用的动态规划解法。公式为dp[j+1]=min(dp[j+1], dp[i]+1) if (s[i] == s[j] && s[i+1] : [j-1]为回文)，判定i：j是不是回文只需要s[i]==s[j]并且s[i+1] : [j-1]是回文就可以了。

dp[ i ]为到 i-1 为止的最小切割数。p控制奇偶数。若判定s[a：b]为回文，再切一刀即dp[a]+1，与dp[b+1]取最小值赋值给dp[b+1]。最后结束的时候多切一刀，因此返回dp
-1