Palindrome Partition I && II

Change-Id: I30f8b3147f6a89739508c58e13c78347a3ae0dc1

Author: applewjg

PalindromePartitioning.java

@@ -0,0 +1,46 @@
+/*
+ Author:     Andy, nkuwjg@gmail.com
+ Date:       Jan 20, 2015
+ Problem:    Palindrome Partitioning
+ Difficulty: Easy
+ Source:     https://oj.leetcode.com/problems/palindrome-partitioning/
+ Notes:
+ 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"]
+ ]
+
+ Solution: ...
+ */
+
+public class Solution {
+    public List<List<String>> partition(String s) {
+        List<List<String>> res = new ArrayList<List<String>>();
+        int n = s.length();
+        boolean[][] dp = new boolean[n][n];
+        for (int i = n - 1; i >= 0; --i) {
+            for (int j = i; j < n; ++j) {
+                dp[i][j]=(s.charAt(i)==s.charAt(j))&&(j<i+2||dp[i+1][j-1]);
+            }
+        }
+        ArrayList<String> path = new ArrayList<String>();
+        dfs(s, dp, 0, path, res);
+        return res;
+    }
+    public void dfs(String s, boolean[][] dp, int start, ArrayList<String> path, List<List<String>> res) {
+        if (s.length() == start) {
+            res.add(new ArrayList<String>(path));
+            return;
+        }
+        for (int i = start; i < s.length(); ++i) {
+            if (dp[start][i] == false) continue;
+            path.add(s.substring(start,i+1));
+            dfs(s, dp, i+1,path,res);
+            path.remove(path.size()-1);
+        }
+    }
+}

PalindromePartitioningII.java

@@ -0,0 +1,32 @@
+/*
+ Author:     Andy, nkuwjg@gmail.com
+ Date:       Jan 23, 2015
+ Problem:    Palindrome Partitioning II
+ Difficulty: Hard
+ Source:     https://oj.leetcode.com/problems/palindrome-partitioning-ii/
+ Notes:
+ Given a string s, partition s such that every substring of the partition is a palindrome.
+ Return the minimum cuts needed for a palindrome partitioning of s.
+ For example, given s = "aab",
+ Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.
+
+ Solution: dp.
+ */
+public class Solution {
+    public int minCut(String s) {
+        int n = s.length();
+        int[] dp = new int[n+1];
+        dp[n] = -1;
+        boolean[][] isP = new boolean[n][n];
+        for (int i = n - 1; i >= 0; --i) {
+            dp[i] = n - 1 - i;
+            for (int j = i; j < n; ++j) {
+                if (s.charAt(i) == s.charAt(j) && (j < i + 2 || isP[i+1][j-1])) isP[i][j] = true;
+                if (isP[i][j] == true) {
+                    dp[i] = Math.min(dp[i], 1 + dp[j+1]);
+                }
+            }
+        }
+        return dp[0];
+    }
+}