Leetcode 131. Palindrome Partitioning

来源：互联网发布：知乎b站三国演义编辑：程序博客网时间：2024/06/11 01:19

DFS non-DP solution.

public class Solution {    public List<List<String>> partition(String s) {        List<List<String>> ret = new ArrayList<>();        List<String> curr = new ArrayList<>();                dfs(0, s, curr, ret);                return ret;    }        private static void dfs(int pos, String s, List<String> curr, List<List<String>> ret) {        if (pos == s.length()) {            ret.add(new ArrayList<>(curr));            return;        }        for (int i=pos; i<s.length(); i++) {            String subStr = s.substring(pos, i+1);            if (isPalindrome(subStr)) {                curr.add(subStr);                dfs(i+1, s, curr, ret);                curr.remove(curr.size()-1);            }        }    }        private static boolean isPalindrome(String s) {        int i=0, j=s.length()-1;        while (i < j) {            if (s.charAt(i) != s.charAt(j)) {                return false;            }            i++;            j--;        }        return true;    }}

Improvement (from 13ms to 7ms). Using a 2D array to save if substring(i, j+1) is a palindrom so that we don't need to check isPalindorme(substring(i, j+1)) every time we call DFS.

public class Solution {    public List<List<String>> partition(String s) {        boolean[][] dp = new boolean[s.length()][s.length()];        buildPalinTab(dp, s);                List<List<String>> ret = new ArrayList<>();        List<String> curr = new ArrayList<>();        dfs(0, s, curr, ret, dp);                return ret;    }        private static void dfs(int pos, String s, List<String> curr, List<List<String>> ret, boolean[][] dp) {        if (pos == s.length()) {            ret.add(new ArrayList<>(curr));            return;        }        for (int i=pos; i<s.length(); i++) {            if (dp[pos][i]) {                String subStr = s.substring(pos, i+1);                curr.add(subStr);                dfs(i+1, s, curr, ret, dp);                curr.remove(curr.size()-1);            }        }    }        private static void buildPalinTab(boolean[][] dp, String s) {        for (int i=s.length()-1; i>=0; i--) {            for (int j=i; j<s.length(); j++) {                if (s.charAt(i) == s.charAt(j)) {                    if (j-i < 3) dp[i][j] = true;                    else dp[i][j] = dp[i+1][j-1];                }            }        }    }}

0 0