Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
Example:
Input: "aab" Output: [ ["aa","b"], ["a","a","b"] ]
Approach #1: backtracking. [C++]
class Solution {
public:
vector<vector<string>> partition(string s) {
vector<vector<string>> ans;
if (s.length() == 0) return ans;
vector<string> path;
dfs(s, 0, ans, path);
return ans;
}
private:
void dfs(const string s, int idx, vector<vector<string>>& ans, vector<string>& path) {
if (s.length() == idx) {
ans.push_back(path);
return;
}
for (int i = idx; i < s.length(); ++i) {
if (isPalindrome(s, idx, i)) {
path.push_back(s.substr(idx, i-idx+1));
// this is the point code in this problem.
dfs(s, i+1, ans, path);
path.pop_back();
}
}
}
bool isPalindrome(const string s, int start, int end) {
while (start <= end) {
if (s[start++] != s[end--])
return false;
}
return true;
}
};
原文：https://www.cnblogs.com/ruruozhenhao/p/10353303.html