【LeetCode】131. Palindrome Partitioning

Palindrome Partitioning

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"]
]

空间换时间。

使用二维数组isPalin记录每个子串是否为回文串。

然后递归来做。可以看做深度优先搜索。

class Solution {
public:
vector<vector<bool> > isPalin;  // isPalin[i][j]==true means s[i,...,j] is palindrome
void buildMap(string s)
{
int n = s.size();
isPalin.resize(n, vector<bool>(n, false));
for(int i = 0; i < n; i ++)
isPalin[i][i] = true;
for(int i = n-1; i >= 0; i --)
{
for(int j = i+1; j < n; j ++)
{
if(s[i] == s[j])
{
if(j == i+1 || isPalin[i+1][j-1] == true)
isPalin[i][j] = true;
}
}
}
}
vector<vector<string>> partition(string s) {
buildMap(s);
vector<vector<string> > ret;
vector<string> cur;
Helper(ret, cur, s, 0);
return ret;
}
void Helper(vector<vector<string> >& ret, vector<string> cur, string s, int offset)
{
if(s == "")
ret.push_back(cur);
for(int i = 0; i < s.size(); i ++)
{
if(isPalin[offset+0][offset+i] == true)
{
cur.push_back(s.substr(0, i+1));    //palin prefix
Helper(ret, cur, s.substr(i+1), offset+i+1);
cur.pop_back();
}
}
}
};