[LeetCode132]Palindrome Partitioning

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#leetcode #DFS #palindrome

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本文介绍了一种使用深度优先搜索(DFS)算法来解决字符串回文划分的问题，通过递归方式找到所有可能的回文子串组合。提供了Java和C++两种实现方案。

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"]
  ]

Analysis:

return all possible palindrome partitioning of s, remind us using DFS.

Java

public class Solution {
    List<List<String>> result;
	List<String> solu;
	public List<List<String>> partition(String s) {
        result = new ArrayList<>();
        solu = new ArrayList<>();
        getPalindrome(s, 0);
        return result;
    }
	public void getPalindrome(String s, int start){
		if(start==s.length()){
			result.add(new ArrayList<>(solu));
			return;
		}
		for(int i=start;i<s.length();i++){
			if(isPalindrome(s.substring(start, i+1))){
				solu.add(s.substring(start, i+1));
				getPalindrome(s, i+1);
				solu.remove(solu.size()-1);
			}
		}
	}
	public boolean isPalindrome(String s){
		if(s.length()<=1) return true;
		int len = s.length();
		for(int i=0;i<len;i++){
			if(s.charAt(i)!=s.charAt(len-1-i))
				return false;
		}
		return true;
	}
}

c++

class Solution {
public:
    bool isPalindrome(string &s, int start, int end){
    while(start < end){
        if(s[start] != s[end])
            return false;
        start++, end--;
    }
    return true;
}
void DFSPalin(vector<vector<string>> &result,
              vector<string> solu,
              string &s,
              int start){
    if(start == s.size()){
        result.push_back(solu);
        return;
    }
    for(int i= start; i<s.size();i++){
        if(isPalindrome(s,start,i)){
            solu.push_back(s.substr(start,i-start+1));
            DFSPalin(result,solu,s,i+1);
            solu.pop_back();
        }
    }
}
vector<vector<string>> partition(string s) {
    vector<vector<string>> result;
    vector<string> solution;
    DFSPalin(result,solution,s,0);
    return result;
}
};