【leetcode】Palindrome Partitioning && Palindrome Partitioning II

Palindrome Partitioning

链接：https://oj.leetcode.com/problems/palindrome-partitioning/

描述：

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

For example, given s =

"aab"

,

Return

[
["aa","b"],
["a","a","b"]
]

解法：
dfs搜索所有解，同时借助dp[i][j]计算从i个字符到第j个字符是否回文。

代码如下：

vector<vector<string>> partition(string s) {
vector<vector<string>> result;
int len  = s.length();
if( len <= 0) return result;
vector<vector<bool> > dp(len, vector<bool>(len, false));
for(int i = 0; i < len; ++i)
dp[i][i] = true;

for(int r = 2; r <= len; ++r)
{
for(int i=0; i <= len - r;++i)
{
int j = i+r-1;
if( s[i] == s[j] && ( i+1 >= j-1 || dp[i+1][j-1]) )
dp[i][j] = true;
}
}

vector<string> path;
dfs(s, 0, dp, path, result);
return result;
}

void dfs(string &s, int start, vector<vector<bool>> &dp,
vector<string> &path, vector<vector<string>> &result)
{
int len = s.length();
if(start == len)
{
result.push_back(path);
return;
}

for(int i=start; i < len; ++i)
{
if( dp[start][i] ){
path.push_back(s.substr(start, i - start + 1));
dfs(s, i+1, dp, path, result);
path.pop_back();
}
}
}

Palindrome
Partitioning II

链接：https://oj.leetcode.com/problems/palindrome-partitioning-ii/

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.

方法一：
两个DP，结果为了图省事，写了一个O(n^3)。超时

代码如下：

int minCut(string s)
{
int len = s.length();
if( len <= 1) return 0;
vector<vector<int>> dp(len, vector<int>(len, INT_MAX));
for(int i=0;i < len; ++i)
dp[i][i] = 0;

for(int r=2; r <= len; ++r)
{
for(int i=0; i <= len - r;++i)
{
int j = i + r - 1;
if(s[i] == s[j] && (i+2 >= j || dp[i+1][j-1] == 0)){
dp[i][j] = 0;
continue;
}
for(int m=i; m < j; ++m)
{
int temp = 1 + dp[i][m] + dp[m+1][j];
if( dp[i][j] > temp)
dp[i][j] = temp;
}
}
}
return dp[0][len-1];
}

方法二：
拆分DP，将第二个DP转化为一位DP。
DP[i] = 0 if flag[0][i] == true
DP[i] = min { DP[j] + flag[j+1][i] ? 1 : (i-j) } 0 <= j < i; else
进一步分析 当flag[j+1][i] 为真时更新就能保证正确性。
代码如下：

int minCut2(string s)
{
int len = s.length();
if(len <= 1) return 0;
vector<vector<bool>> dp(len, vector<bool>(len, false));
for(int i=0;i < len; ++i)
dp[i][i] = true;

for(int r=2; r <= len; ++r)
{
for(int i=0; i <= len-r; ++i)
{
int j = r + i - 1;
if( s[i] == s[j] && (i + 2 >= j || dp[i+1][j -1]))
dp[i][j] = true;
}
}

vector<int> output(len, 0);
for(int i=1;i < len; ++i)
{
if( dp[0][i] ){
output[i] = 0;
continue;
}else{
output[i] = i;
}
for(int j = i; j > 0; --j)
{
if( dp[j][i] )
output[i] = min(output[i], output[j-1] + 1);
}
}
return output[len-1];
}