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【LeetCode】Palindrome Partitioning
Palindrome Partitioning
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"] ]
空间换时间。
存放每个子串是否为回文串(O(n2))
然后递归来做。可以看做深度优先搜索。
class Solution {public: bool isPal(string substr) { for(string::size_type i = 0, j = substr.size()-1; i < j; i++, j--) { if(substr[i] != substr[j]) return false; } return true; } vector<vector<string>> partition(string s) { map<string, bool> m; //space for time for(string::size_type i = 0; i < s.size(); i ++) { for(string::size_type j = i; j < s.size(); j ++) { string sub = s.substr(i, j-i+1); if(isPal(sub)) m[sub] = true; else m[sub] = false; } } vector<string> v; vector<vector<string> > ret; Helper(s, v, ret, m); return ret; } void Helper(string sub, vector<string> v, vector<vector<string>>& ret, map<string, bool>& m) { for(string::size_type st = 0; st < sub.size(); st ++) { if(m[sub.substr(0, st+1)]) { v.push_back(sub.substr(0,st+1)); if(st == sub.size()-1) ret.push_back(v); else { //recursion Helper(sub.substr(st+1, sub.size()-1-st), v, ret, m); v.pop_back(); } } } }};
判断回文子串的时候可以用DP的思想，若s[i,j]为回文串，并且s[i-1]==s[j+1]，则s[i-1,j+1]也为回文串。
class Solution {public: vector<vector<string>> partition(string s) { map<string, bool> m; //space for time for(int i = s.size()-1; i >= 0; i --) {//string::size_type will overflow for(int j = i; j < s.size(); j ++) { m[s.substr(i,j-i+1)] = false; if(i == j) m[s.substr(i,1)] = true; else { if(s[i] == s[j]) { //s[i,j] or s[i+1, j-1] if(j == i+1 || m[s.substr(i+1, j-i-1)]) m[s.substr(i,j-i+1)] = true; } } } } vector<string> v; vector<vector<string> > ret; Helper(s, v, ret, m); return ret; } void Helper(string sub, vector<string> v, vector<vector<string>>& ret, map<string, bool>& m) { if(m[sub]) { v.push_back(sub); ret.push_back(v); v.pop_back(); } //not else for(string::size_type st = 0; st < sub.size()-1; st ++) { if(m[sub.substr(0, st+1)]) { v.push_back(sub.substr(0,st+1)); //recursion Helper(sub.substr(st+1, sub.size()-1-st), v, ret, m); v.pop_back(); } } }};
【LeetCode】Palindrome Partitioning
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