leetcode Palindrome Partitioning

Description:

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"]]

Solution:

1.DFS

在每一步都可以判断中间结果是否为合法结果，用回溯法。 一个长度为n的字符串，有 n -1 个地方可以隔断，每个地方都有两个情况，因此复杂度为O（2^（n-1）

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

2.DP

res(i)表示s[0...i-1]的所有分解方式
isPalin(i,j)表示s[i...j]是否为回文串
isPalin(i,j) = true  if  i==j or (s[i] == s[j] and isPalin(i + 1, j - 1)) or (s[i] == s[j] and i + 1 == j)
res(i) = res(j) + s[j...i-1]  if isPalin(j, i-1)

class Solution {public:    vector<vector<string>> partition(string s) {        int size = s.size();        vector<vector<vector<string>>> result(size+1);        result[0].push_back(vector<string>(0));        vector<vector<bool>> isPalin(size + 1, vector<bool>(size + 1, false));                for (int i = 0; i < size; i++) {            for (int j = 0; j <= i; j++) {                if (s[i] == s[j] && (i - j < 2 || isPalin[j + 1][i - 1])) {                    isPalin[j][i] = true;                    for (int k = 0; k < result[j].size(); k++) {                        vector<string> prefix = result[j][k];                        prefix.push_back(s.substr(j,i-j+1));                        result[i+1].push_back(prefix);//re(i) = re(j) + s[j...i-1]  if isPalin(j, i-1)                    }                }            }        }        return result[size];    }};