[LeetCode] Palindrome Partitioning

最新推荐文章于 2019-09-02 10:48:14 发布

原创最新推荐文章于 2019-09-02 10:48:14 发布 · 298 阅读

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本文介绍了一种使用深度搜索加回溯的方法来解决字符串的所有可能回文分割问题。通过递归地检查子串是否为回文，并将符合条件的子串添加到当前分割方案中，最终收集所有有效的分割组合。

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"]
  ]

深搜加回溯，从i到j如果是回文，将s[i,j]加入临时切分结果，继续判断从j开始的回文分割。

class Solution {
public:
    vector<vector<string>> partition(string s) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        vector<vector<string>> result;
        vector<string> temp;
        if(s.length() == 1) {
            temp.push_back(s);
            result.push_back(temp);
            return result;
        }
        recursivePair(s,0,temp,result);
        return result;
    }

    void recursivePair(string s, int start, vector<string> temp, vector<vector<string>> &result) {
        if(start == s.length()) {
            result.push_back(temp);
        }    
        for(int i = start; i < s.length(); i++) {
            if(isPalindrome(s,start,i)) {
                temp.push_back(s.substr(start,i+1-start));
                recursivePair(s,i+1,temp,result);
                temp.pop_back();
            }
        }
    }   
    
    bool isPalindrome(string s,int start, int end) {
        if(start == end) return true;
        for(int i = start, j = end; i < j; i++,j--) {
            if(s[i] != s[j]) return false;
        }
        return true;
    }
};