【leetcode】131. Palindrome Partitioning【java】

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本文介绍了一种递归算法，用于将输入字符串分割成所有可能的子串组合，确保每个子串都是回文串。通过回溯方法实现，具体步骤包括检查子串是否为回文及收集所有符合条件的分割方案。

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"]
]

public class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> res = new ArrayList<List<String>>();
        List<String> list = new ArrayList<String>();
        backTrack(s, 0, list, res);
        return res;
    }
    public void backTrack(String s, int pos, List<String> list, List<List<String>> res) {
        if (pos == s.length()) {
            res.add(new ArrayList<String>(list));
        }
        for (int i = pos; i < s.length(); i++) {
            if (isPal(s, pos, i)) {
                list.add(s.substring(pos, i + 1));
                backTrack(s, i + 1, list, res);
                list.remove(list.size() - 1);
            }
        }
    }
    public boolean isPal(String s, int low, int high) {
        while (low < high) {
            if (s.charAt(low++) != s.charAt(high--)) {
                return false;
            }
        }
        return true;
    }
}