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"]
]
题意：找出一个字符串所有可能的回文子串组合。
思路：深度搜索所有组合，每找出一个子串，就判断一下是否是回文串，找到最后一个位置，那么就是一个符合条件的组合。
public List<List<String>> partition(String s) {
List<List<String>> res = new ArrayList<>();
if (s == null || s.length() == 0) {
return res;
}
HashMap<String, Boolean> memory = new HashMap<>();
dfsPalindrome(s, 0, new ArrayList<>(), res, memory);
return res;
}
private void dfsPalindrome(String s, int start, List<String> subres, List<List<String>> res, HashMap<String, Boolean> memory) {
if (start == s.length()) {
res.add(new ArrayList<>(subres));
return;
}
for (int end = start + 1; end <= s.length(); end++) {
String tmp = s.substring(start, end);
if (!isPalindrome(tmp, memory)) {
continue;
}
subres.add(tmp);
dfsPalindrome(s, end, subres, res, memory);
subres.remove(subres.size() - 1);
}
}
private boolean isPalindrome(String tmp, HashMap<String, Boolean> memory) {
if (memory.containsKey(tmp)) {
return true;
}
int start = 0, end = tmp.length() - 1;
while (start < end) {
if (tmp.charAt(start) != tmp.charAt(end)) {
return false;
}
start++;
end--;
}
memory.put(tmp, true);
return true;
}