lintcode136- Palindrome Partitioning- medium

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

Example

Given s = "aab", return:

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

DFS。想成每次从允许的index开始，for循环试试切割到某个位置拿出字符串来。如果切出不是回文的肯定不合格，放弃。如果切出的是回文的，那递归检查后面的substring们。如果切出来的正好切完最后一个字母了，那之前切法就是合格的，出口了放入result。

这里有一个可以优化的地方，就是提前把boolean[][] isPalindrome 数组准备好，之后就可以在重复寻求判断的时候O(1)而不是O(string长度)拿到结果。进一步优化创造boolean[][]数组的方法是动态规划，isPalindrome[i][j] = isPalindrome[i+1][j-1] && s.charAt(i) == s.charAt(j)，每个

1. 普通准备boolean[][]数字

public class Solution {
    /*
     * @param s: A string
     * @return: A list of lists of string
     */
    public List<List<String>> partition(String s) {
        // write your code here
        List<List<String>> result = new ArrayList<>();
        if (s == null) {
            return result;
        }
        if (s.length() == 0) {
            List<String> list = new ArrayList<>();
            list.add("");
            result.add(list);
            return result;
        }
        
        boolean[][] isPalindromes = new boolean[s.length()][s.length()];
        for (int i = 0; i < s.length(); i++) {
            for (int j = i; j < s.length(); j++) {
                isPalindromes[i][j] = isPalindrome(s.substring(i, j+1));
            }
        }
        
        dfs(s, 0, isPalindromes, new ArrayList<String>(), result);
        return result;
    }
    
    private void dfs(String s, int idx, boolean[][] isPalindromes, List<String> crt, List<List<String>> result) {
        
        if (idx == s.length()) {
            result.add(new ArrayList<String>(crt));
            return;
        }
        
        for (int i = idx; i < s.length(); i++) {
            if (!isPalindromes[idx][i]) {
                continue;
            }
            crt.add(s.substring(idx, i + 1));
            dfs(s, i + 1, isPalindromes, crt, result);
            crt.remove(crt.size() - 1);
        }
    }
    
    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.动态规划准备boolean[][]数组（最好纸上画一下ij的矩阵，来判断下要利用boolean[i+1][j-1]的话要先准备哪些数据，之后要按什么顺序来算。实际上是先准备对角线[i][i]一条，对角线右侧[i][i+1]一条，之后要让i从下往上，j从左到右来算。）

public class Solution {
    /*
     * @param s: A string
     * @return: A list of lists of string
     */
    public List<List<String>> partition(String s) {
        // write your code here
        List<List<String>> result = new ArrayList<>();
        if (s == null) {
            return result;
        }
        if (s.length() == 0) {
            List<String> list = new ArrayList<>();
            list.add("");
            result.add(list);
            return result;
        }
        
        boolean[][] isPalindromes = isPalindromes(s);
        dfs(s, 0, isPalindromes, new ArrayList<String>(), result);
        return result;
    }
    
    private void dfs(String s, int idx, boolean[][] isPalindromes, List<String> crt, List<List<String>> result) {
        
        if (idx == s.length()) {
            result.add(new ArrayList<String>(crt));
            return;
        }
        
        for (int i = idx; i < s.length(); i++) {
            if (!isPalindromes[idx][i]) {
                continue;
            }
            crt.add(s.substring(idx, i + 1));
            dfs(s, i + 1, isPalindromes, crt, result);
            crt.remove(crt.size() - 1);
        }
    }
    
    private boolean[][] isPalindromes(String s) {
        boolean[][] isPalindromes = new boolean[s.length()][s.length()];
        
        for (int i = 0; i < s.length(); i++) {
            isPalindromes[i][i] = true;
        }
        for (int i = 0; i < s.length() - 1; i++) {
            isPalindromes[i][i + 1] = s.charAt(i) == s.charAt(i + 1);
        }
        for (int i = s.length() - 3; i >= 0; i--) {
            for (int j = i + 2; j < s.length(); j++) {
                isPalindromes[i][j] = isPalindromes[i + 1][j - 1] && s.charAt(i) == s.charAt(j);
            }
        }
        return isPalindromes;
    }
}