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"]

链接:  http://leetcode.com/problems/palindrome-partitioning/

题解:

一看到return all xxxx,就猜到可能要用回溯。这道题就是比较典型的递归+回溯。递归前要判断当前的子字符串是否palindrome,答案是false的话要continue。

Time Complexity - O(n*2n), Space Complexity - O(n*2n)

public class Solution {
public List<List<String>> partition(String s) {
List<List<String>> res = new ArrayList<>();
if(s == null || s.length() == 0)
return res;
ArrayList<String> list = new ArrayList<>();
partition(res, list, s, 0);
return res;
} private void partition(List<List<String>> res, ArrayList<String> list, String s, int pos) {
if(pos == s.length()) {
res.add(new ArrayList<String>(list));
return;
} for(int i = pos + 1; i <= s.length(); i++) {
String partition = s.substring(pos, i);
if(!isPalindrome(partition))
continue;
list.add(partition);
partition(res, list, s, i);
list.remove(list.size() - 1);
}
} private boolean isPalindrome(String s) {
int lo = 0, hi = s.length() - 1; while(lo < hi) {
if(s.charAt(lo) != s.charAt(hi))
return false;
lo++;
hi--;
} return true;
}
}

需要好好看看主方法来确定定量分析递归算法的时间复杂度。

二刷:

仔细想一想代码可以简化不少。主要分为三部分。1是题目给定的方法,2是辅助方法,用来递归和回溯,3是判断string是否是palindrome。注意考虑清楚需要多少变量,以及时间空间复杂度。

Time Complexity: O(n!)

Space Complexity: O(n ^ 2)

Java:

public class Solution {
public List<List<String>> partition(String s) {
List<List<String>> res = new ArrayList<>();
List<String> list = new ArrayList<>();
partition(res, list, s);
return res;
} private void partition(List<List<String>> res, List<String> list, String s) {
if (s == null || s.length() == 0) {
res.add(new ArrayList<String>(list));
return;
}
for (int i = 0; i < s.length(); i++) {
String subStr = s.substring(0, i + 1);
if (isPalindrome(subStr)) {
list.add(subStr);
partition(res, list, s.substring(i + 1));
list.remove(list.size() - 1);
}
}
} private boolean isPalindrome(String s) {
if (s == null || s.length() < 2) {
return true;
}
int lo = 0, hi = s.length() - 1;
while (lo <= hi) {
if (s.charAt(lo) != s.charAt(hi)) {
return false;
}
lo++;
hi--;
}
return true;
}
}

三刷:

依然是使用二刷的方法。

Java:

public class Solution {
public List<List<String>> partition(String s) {
List<List<String>> res = new ArrayList<>();
if (s == null || s.length() == 0) return res;
partition(res, new ArrayList<>(), s);
return res;
} private void partition(List<List<String>> res, List<String> list, String s) {
if (s.length() == 0) {
res.add(new ArrayList<String>(list));
return;
}
for (int i = 0; i <= s.length(); i++) {
String front = s.substring(0, i);
if (isPalindrome(front)) {
list.add(front);
partition(res, list, s.substring(i));
list.remove(list.size() - 1);
}
}
} private boolean isPalindrome(String s) {
if (s == null || s.length() == 0) return false;
int lo = 0, hi = s.length() - 1;
while (lo < hi) {
if (s.charAt(lo) != s.charAt(hi)) return false;
lo++;
hi--;
}
return true;
}
}

Reference:

http://stackoverflow.com/questions/24591616/whats-the-time-complexity-of-this-algorithm-for-palindrome-partitioning

http://blog.csdn.net/metasearch/article/details/4428865

https://en.wikipedia.org/wiki/Master_theorem

http://www.cnblogs.com/zhuli19901106/p/3570430.html

https://leetcode.com/discuss/18984/java-backtracking-solution

https://leetcode.com/discuss/9623/my-java-dp-only-solution-without-recursion-o-n-2

https://leetcode.com/discuss/41626/concise-java-solution

https://leetcode.com/discuss/4788/shouldnt-we-use-dp-in-addition-to-dfs

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