131. Palindrome Partitioning

Given a string s, partition s such that every substring of the partition is a palindrome.
Return all possible palindrome partitioning of s.
Example:

Input: "aab"
Output:
[
  ["aa","b"],
  ["a","a","b"]
]

难度：medium

题目：给定一字符串，切分字符串使得每个子串为回文字符串。返回所有可能的切分。

思路：
首先利用动态规划找出所有回文
定义pi 表示从子串p从i到j是回文。
转换方程
p[i][j] = p[i + 1][j - 1] && s[i] == s[j]
然后利用递归收集所有可能的切分

Runtime: 3 ms, faster than 99.87% of Java online submissions for Palindrome Partitioning.
Memory Usage: 39.5 MB, less than 100.00% of Java online submissions for Palindrome Partitioning.

class Solution {
    public List<List<String>> partition(String s) {
        if (null == s || s.length() < 1) {
            return new ArrayList();
        }
        int n = s.length();
        if (1 == s.length()) {
            return Arrays.asList(Arrays.asList(s));
        }
        
        char[] sc = s.toCharArray();
        boolean[][] bc = new boolean[n][n];
        // 初始化 p1 p2
        for (int i = 0; i < n - 1; i++) {
            bc[i][i] = true;
            if (sc[i] == sc[i + 1]) {
                bc[i][i + 1] = true;
            }
        }
        bc[n - 1][n - 1] = true;
        // i indicates string length
        for (int i = 3; i <= n; i++) {
            // j indicates string index
            for (int j = 0; j <= n - i; j++) {
                boolean b = (sc[j] == sc[j + i - 1]);
                bc[j][j + i - 1] = b && bc[j + 1][j + i - 1 - 1];
            }
        }
        //收集切分
        List<List<String>> result = new ArrayList<>();
        buildPalindrome(bc, result, new Stack<>(), s, 0, n);
        
        return result;
    }
    
    private void buildPalindrome(boolean[][] bc,
        List<List<String>> result, Stack<String> stack, 
        String s, int depth, int n) {
        if (n == depth) {
            result.add(new ArrayList(stack));
            return;
        }

        for (int i = depth; i < n; i++) {
            if (bc[depth][i]) {
                stack.push(s.substring(depth, i + 1));
                buildPalindrome(bc, result, stack, s, i + 1, n);
                stack.pop();
            }
        }
    }
}