Palindrome Partitioning - Leetcode

来源：互联网发布：诺基亚5320手电筒软件编辑：程序博客网时间：2024/06/10 04:47

思路：在每一步，一一往前搜索并验证是否是回文，如果当前不是回文则停止前进，返回上一个状态并记录。每一次都搜索到结尾，因为这里要累计回文子串。回溯法就可以用到，在这里。

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"]  ]

DP 方法：这里我用哈希表记录了每一个index-结尾的值。

public class Solution {    public List<List<String>> partition(String s) {       boolean[][] f = new boolean[s.length()][s.length()];for (int i = f.length - 1; i >= 0; i--) {for (int j = i; j < f.length; j++) {if (s.charAt(i) == s.charAt(j)&& (j - i < 2 || f[i + 1][j - 1] == true))f[i][j] = true;}}HashMap<Integer, List<List<String>>> hm = new HashMap<>();for (int i = f.length - 1; i >= 0; i--) {for (int j = i; j < f.length; j++) {if (f[i][j]) {String parlindrome = s.substring(i, j + 1);if (j < f.length - 1) {for (List<String> kks_list : hm.get(j + 1)) {List<String> ls = new ArrayList<>(kks_list);ls.add(0, parlindrome);if (!hm.containsKey(i)) {List<List<String>> kks = new ArrayList<>();kks.add(ls);hm.put(i, kks);} elsehm.get(i).add(ls);}} else {ArrayList<String> as = new ArrayList<String>();as.add(parlindrome);List<List<String>> lls = new ArrayList<>();lls.add(as);if (!hm.containsKey(i))hm.put(i, lls);elsehm.get(i).add(as);}}}}return hm.get(0);    }}

深搜DFS：这个思路清楚就清楚了，复杂度2^n

public class Solution {    public List<List<String>> partition(String s) {       List<List<String>> result = new ArrayList<>();       List<String> path = new ArrayList<>();       DFS(s, result, 0, path);       return result;    }    private void DFS(String s, List<List<String>> result, int start, List<String> path){        if(start == s.length()){            result.add(<span style="background-color: rgb(255, 255, 51);">new ArrayList<String>(path)</span>);            return;        }        for(int i=start; i<s.length(); i++){            if(isPalindrome(s,start,i)){ // cut from i            String sub = s.substring(start, i+1);                path.add(sub);                DFS(s, result, i+1, path); //continue cutting                path.remove(path.size()-1);            }        }    }    private boolean isPalindrome(String s, int start, int end){        while(s.charAt(start) == s.charAt(end) && start<end)        {            start++;            end--;        }        return start>=end;    }}

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