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LeetCode-131-Palindrome-Partitioning.java
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LeetCode-131-Palindrome-Partitioning.java
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/*
LeetCode: https://leetcode.com/problems/palindrome-partitioning/
LintCode: http://www.lintcode.com/problem/palindrome-partitioning/
JiuZhang: http://www.jiuzhang.com/solutions/palindrome-partitioning/
ProgramCreek: http://www.programcreek.com/2013/03/leetcode-palindrome-partitioning-java/
Analysis:
1.Pure DFS
Time: O(N*2^N)
2.DFS + DP
Time: O(2^N)
https://leetcode.com/problems/palindrome-partitioning/discuss/41982/Java-DP-%2B-DFS-solution
3. Pure DP
Time: O(2^N), in worst case, all chars are the same, so total 2^(N-1) combination, and every step copy the whole result set again, so the worst time complexity is O(2^N).
https://leetcode.com/problems/palindrome-partitioning/discuss/41974/My-Java-DP-only-solution-without-recursion.-O(n2)
*/
public class Solution {
// 1.Pure DFS
// public List<List<String>> partition(String s) {
// List<List<String>> result = new ArrayList<List<String>>();
// List<String> list = new ArrayList<String>();
// if(s == null || s.length() == 0) return result;
// DFS(s, 0, result, list);
// return result;
// }
// private void DFS(String s, int start, List<List<String>> result, List<String> list){
// // stop condition
// if(start == s.length()){
// result.add(new ArrayList<String>(list));
// return;
// }
// for(int end = start; end < s.length(); end++){
// if(isPalindrome(s, start, end)){
// list.add(s.substring(start, end + 1));
// DFS(s, end + 1, result, list);
// list.remove(list.size() - 1);
// }
// }
// }
// private boolean isPalindrome(String s, int lo, int hi) {
// while (lo < hi) {
// if (s.charAt(lo) != s.charAt(hi)) return false;
// lo++;
// hi--;
// }
// return true;
// }
// 2. DFS + DP (Memorized DFS)
// public List<List<String>> partition(String s) {
// List<List<String>> result = new ArrayList<List<String>>();
// List<String> list = new ArrayList<String>();
// boolean[][] dp = new boolean[s.length()][s.length()];
// if(s == null || s.length() == 0) return result;
// helper(s, 0, dp, result, list);
// return result;
// }
// private void helper(String s, int start, boolean[][] dp, List<List<String>> result, List<String> list) {
// if (start == s.length()) {
// result.add(new ArrayList<>(list));
// return;
// }
// for (int end = start; end < s.length(); end++) {
// // "abca", start = 0, end = 3, so "a"=="a" -> ok
// if (s.charAt(start) != s.charAt(end)) continue;
// // "abca", start + 1 = 1, end - 1= 2, so "b"!="c" -> continue
// if (start + 1 < end - 1 && !dp[start + 1][end - 1]) continue;
// dp[start][end] = true;
// list.add(s.substring(start, end + 1));
// helper(s, end + 1, dp, result, list);
// list.remove(list.size() - 1);
// }
// }
// 3. Pure DP
public List<List<String>> partition(String s) {
if(s == null || s.length() == 0) return new ArrayList<>();
int len = s.length();
List<List<String>>[] result = new List[len + 1]; // dp results from 0 length to actual length
result[0] = new ArrayList<>();
result[0].add(new ArrayList<>());
boolean[][] dp = new boolean[len][len];
for (int r = 0; r < len; r++) {
result[r + 1] = new ArrayList<>();
for (int l = 0; l <= r; l++) {
// First approach
// if (s.charAt(l) != s.charAt(r)) continue;
// if(l + 1 < r - 1 && !dp[l + 1][r - 1]) continue;
// dp[l][r] = true;
// String sSub = s.substring(l, r + 1);
// for(List<String> sub : result[l]) {
// List<String> temp = new ArrayList<>(sub);
// temp.add(sSub);
// result[r + 1].add(temp);
// }
// Second approach
if (s.charAt(l) != s.charAt(r)) continue;
if(l + 1 < r - 1 && !dp[l + 1][r - 1]) continue;
if (s.charAt(l) == s.charAt(r) && (r - l <= 2 || dp[l + 1][r - 1])) {
dp[l][r] = true;
String sSub = s.substring(l, r + 1);
for(List<String> sub : result[l]) {
List<String> temp = new ArrayList<>(sub);
temp.add(sSub);
result[r + 1].add(temp);
}
}
}
}
return result[len];
}
}