136.分割回文串

描述

给定一个字符串s，将s分割成一些子串，使每个子串都是回文串。
返回s所有可能的回文串分割方案。

样例

给出 s = "aab"，返回

[
  ["aa", "b"],
  ["a", "a", "b"]
]

空集应该算回文

演示

示例

代码

1. shorter but slower

public class Solution {
    /**
     * @param s: A string
     * @return: A list of lists of string
     */
    public List> partition(String s) {
        List> results = new ArrayList<>();
        if (s == null || s.length() == 0) {
            // return results表示切割方案为空，对于输入空集也适用
            return results;
        }
        List partition = new ArrayList();
        helper(s, 0, partition, results);
        return results;
    }
    
    private void helper(String s,
                        int startIndex,
                        List partition,
                        List> results) {
        if (startIndex == s.length()) {
            results.add(new ArrayList(partition));
            return;
        }
        
        for (int i = startIndex; i < s.length(); i++) {
            String subString = s.substring(startIndex, i + 1);
            // 如果切割的字符串不是回文代表当前 i 位置切割的方案不可行
            if (!isPalindrome(subString)) {
                continue;
            }
            partition.add(subString);
            helper(s, i + 1, partition, results);
            partition.remove(partition.size() - 1);
        }
    }

    // 记住回文判断的写法
    private boolean isPalindrome(String s) {
        for (int i = 0, j = s.length() - 1; i < j; i++, j--) {
            if (s.charAt(i) != s.charAt(j)) {
                return false;
            }
        }
        return true;
    }
}

2. longer but faster

public class Solution {
    List> results;
    boolean[][] isPalindrome;
    
    /**
     * @param s: A string
     * @return: A list of lists of string
     */
    public List> partition(String s) {
        results = new ArrayList<>();
        if (s == null || s.length() == 0) {
            return results;
        }
        
        getIsPalindrome(s);
        
        helper(s, 0, new ArrayList());
        
        return results;
    }
    
    private void getIsPalindrome(String s) {
        int n = s.length();
        isPalindrome = new boolean[n][n];
        
        for (int i = 0; i < n; i++) {
            isPalindrome[i][i] = true;
        }
        for (int i = 0; i < n - 1; i++) {
            isPalindrome[i][i + 1] = (s.charAt(i) == s.charAt(i + 1));
        }
        
        for (int i = n - 3; i >= 0; i--) {
            for (int j = i + 2; j < n; j++) {
                isPalindrome[i][j] = isPalindrome[i + 1][j - 1] && s.charAt(i) == s.charAt(j);
            }
        }
    }
    
    private void helper(String s,
                        int startIndex,
                        List partition) {
        if (startIndex == s.length()) {
            addResult(s, partition);
            return;
        }
        
        for (int i = startIndex; i < s.length(); i++) {
            if (!isPalindrome[startIndex][i]) {
                continue;
            }
            partition.add(i);
            helper(s, i + 1, partition);
            partition.remove(partition.size() - 1);
        }
    }
    
    private void addResult(String s, List partition) {
        List result = new ArrayList<>();
        int startIndex = 0;
        for (int i = 0; i < partition.size(); i++) {
            result.add(s.substring(startIndex, partition.get(i) + 1));
            startIndex = partition.get(i) + 1;
        }
        results.add(result);
    }
}