Lintcode 136. Split palindrome
Title description: Given a string, split it, and each part after splitting is a palindrome string (these split parts can form the original string). Finally, we return to all possible such palindrome splitting schemes.
One conclusion can be remembered: all cutting problems are combined problems.
Why do you say that? For example, to cut "abc", you can put a number in the middle of them to become: a1b2c, so the corresponding relationship between cutting and combination is as follows:
Cutting plan: a | b | c Corresponding combination plan: [1, 2]
Cutting plan: a | bc Corresponding combination plan: [1]
Cutting plan: ab | c Corresponding combination plan: [2]
Cutting plan: abc Corresponding combination plan: []
So: the problem of cutting n letters ==> the problem of combining n-1 numbers
The code implementation uses DFS to imitate the problem of combination to solve:
class Solution {
public:
/*
* @param s: A string
* @return: A list of lists of string
*/
vector<vector<string>> partition(string &s) {
vector<vector<string>> result;
if (0 == s.length()) {
return result;
}
vector<string> partition;
//4、递归的调用
helper(s, 0, partition, result);
return result;
}
//1、递归的定义
void helper(string &s, int startIdx, vector<string> &partition, vector<vector<string>> &result) {
//3、递归的出口
if (s.size() == startIdx) {
result.push_back(partition);
return;
}
//2、递归的拆解
//"aaab"
for (int i = startIdx; i < s.size(); ++i) {
string substring = s.substr(startIdx, i - startIdx + 1);
if (!isPalindrome(substring)) {
continue;
}
partition.push_back(substring);
helper(s, i + 1, partition, result);
partition.pop_back();
}
}
//回文串判断
bool isPalindrome(string &s) {
for (int i = 0, j = s.size() - 1; i < j; ++i, --j) {
if (s.at(i) != s.at(j)) {
return false;
}
}
return true;
}
};