Leetcode--Palindrome Partitioning II
2014-09-11 20:06
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Problem Description:
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 =
Return
be produced using 1 cut.
分析:设cut[i] = 区间[0,i]之间最小的cut数,n为字符串长度, 则,
cut[i] = min(cut[i],1+cut[j] ) 0<=j<i
有个转移函数之后,一个问题出现了,就是如何判断[j,i]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
flag[i][j] = true if [i,j]为回文
那么
flag[i][j] = str[i] == str[j] && P[i+1][j-1];
class Solution {
public:
int minCut(string s) {
//if(s.size()==0)
// return 0;
int n=s.size();
vector<vector<int> > flag(n,vector<int>(n,0));
vector<int> cut(n+1);
for(int i=0;i<=n;i++)
cut[i]=i-1;
for(int i=0;i<n;i++)
{
for(int j=0;j<=i;j++)
{
if(s[i]==s[j]&&(i-j<2||flag[j+1][i-1]==1))
{
flag[j][i]=1;
cut[i+1]=min(cut[i+1],cut[j]+1);
}
}
}
return cut
;
}
};
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
1since the palindrome partitioning
["aa","b"]could
be produced using 1 cut.
分析:设cut[i] = 区间[0,i]之间最小的cut数,n为字符串长度, 则,
cut[i] = min(cut[i],1+cut[j] ) 0<=j<i
有个转移函数之后,一个问题出现了,就是如何判断[j,i]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
flag[i][j] = true if [i,j]为回文
那么
flag[i][j] = str[i] == str[j] && P[i+1][j-1];
class Solution {
public:
int minCut(string s) {
//if(s.size()==0)
// return 0;
int n=s.size();
vector<vector<int> > flag(n,vector<int>(n,0));
vector<int> cut(n+1);
for(int i=0;i<=n;i++)
cut[i]=i-1;
for(int i=0;i<n;i++)
{
for(int j=0;j<=i;j++)
{
if(s[i]==s[j]&&(i-j<2||flag[j+1][i-1]==1))
{
flag[j][i]=1;
cut[i+1]=min(cut[i+1],cut[j]+1);
}
}
}
return cut
;
}
};
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