Palindrome Partitioning

题目：

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"]
]

算法思想：

1，递归写法：从首字符k开始，找到字符串s中下个k，判断以两字符为首尾的子字符串A是否为回文，如果是的话用余下的字符串B作为参数传递给递归函数，然后将A插入到递归函数返回的结果中；如果不是的话继续寻找直至到字符串末尾。

2，迭代写法：这个可以在构建出IsPal[][]的基础上得到，参见点击打开链接。

vector> partition(string s)
{
vector> strss;

if (s.length() == 0)
return strss;
else if (s.length() == 1)
{
strss.push_back(vector(1, s));
return strss;
}
else
{
char c = s[0];

for (int i = 0; i < s.length(); i++)
{
i = s.find_first_of(c, i);

if (i == -1) break;

string s_temp1 = s.substr(0, i+1);
string s_temp2(s_temp1.rbegin(), s_temp1.rend());

if (s_temp1 == s_temp2)
{
vector> strss_temp = partition(s.substr(i+1, s.length()-1));

if (strss_temp.size() == 0)
{
strss.push_back(vector(1, s_temp1));
}
else
{
for (int j = 0; j < strss_temp.size(); j++)
{
strss_temp[j].insert(strss_temp[j].begin(), s_temp1);
strss.push_back(strss_temp[j]);
}
}
}
}

return strss;
}

}