Leetcode: Palindrome Partition I II

 

题目一, 题目二

 

 

思路

1. 第一遍做时就参考别人的, 现在又忘记了 做的时候使用的是二维动态规划, 超时加超内存

 

2. 只当 string 左部分是回文的时候才有可能减少 cut

 

3. 一维动规. 令 cuts[i] 表示string[i, string.size()] 所需的切割数, 那么

状态转移方程为 cuts[i] = min(cuts[j]+1) j > i && string[i, j] is palindrome

时间复杂度上仍是 o(n*n), 但更新 cuts 的限制条件比较多了, cuts[i] 更新频率较低 

 

代码:

超时二维动规代码

#include <iostream>
#include <memory.h>
using namespace std;int cuts[1000][1000];
int palindrom[1000][1000];
const int INFS = 0x3f3f3f3f;
class Solution {
public:int minCut(string s) {memset(cuts, 0x3f, sizeof(cuts));memset(palindrom, 0x3f, sizeof(palindrom));int curcuts = countCuts(s,0,s.size()-1);return curcuts;}int countCuts(string &s, int i, int j) {if(j <= i) return 0;if(isPalindrome(s,i,j))return (cuts[i][j]=0);if(cuts[i][j] != INFS)return cuts[i][j];int curcuts = INFS;for(int k = i; k < j; k++) {curcuts = min(curcuts, 1+countCuts(s,i,k)+countCuts(s,k+1,j));}return (cuts[i][j]=curcuts);}bool isPalindrome(string &s, int i, int j) {if(palindrom[i][j] == 1)return true;if(j <= i)return (palindrom[i][j] = true);if(palindrom[i][j] == 0)return false;return (palindrom[i][j] = (s[i]==s[j] && isPalindrome(s,i+1,j-1)));}
};int main() {string str = "apjesgpsxoeiokmqmfgvjslcjukbqxpsobyhjpbgdfruqdkeiszrlmtwgfxyfostpqczidfljwfbbrflkgdvtytbgqalguewnhvvmcgxboycffopmtmhtfizxkmeftcucxpobxmelmjtuzigsxnncxpaibgpuijwhankxbplpyejxmrrjgeoevqozwdtgospohznkoyzocjlracchjqnggbfeebmuvbicbvmpuleywrpzwsihivnrwtxcukwplgtobhgxukwrdlszfaiqxwjvrgxnsveedxseeyeykarqnjrtlaliyudpacctzizcftjlunlgnfwcqqxcqikocqffsjyurzwysfjmswvhbrmshjuzsgpwyubtfbnwajuvrfhlccvfwhxfqthkcwhatktymgxostjlztwdxritygbrbibdgkezvzajizxasjnrcjwzdfvdnwwqeyumkamhzoqhnqjfzwzbixclcxqrtniznemxeahfozp";cout << str.size() << endl;cout << (new Solution())->minCut(str) << endl;return 0;
}

　　

优化后的一维动规

#include <iostream>
#include <memory.h>
using namespace std;int cuts[1500];
int palindrom[1500][1500];
const int INFS = 0x3f3f3f3f;
class Solution {
public:int minCut(string s) {memset(cuts, 0x3f, sizeof(cuts));memset(palindrom, 0x3f, sizeof(palindrom));int curcuts = countCuts(s,0,s.size()-1);return curcuts;}int countCuts(string &s, int i, int j) {if(j <= i) return 0;if(isPalindrome(s,i,j))return 0;if(cuts[i] != INFS)return cuts[i];int curcuts = INFS;for(int k = i; k < j; k++) {if(isPalindrome(s,i,k))curcuts = min(curcuts, 1+countCuts(s,k+1,j));}return (cuts[i]=curcuts);}bool isPalindrome(string &s, int i, int j) {if(palindrom[i][j] == 1)return true;if(j <= i)return (palindrom[i][j] = true);if(palindrom[i][j] == 0)return false;return (palindrom[i][j] = (s[i]==s[j] && isPalindrome(s,i+1,j-1)));}
};int main() {string str = "bb";cout << str.size() << endl;cout << (new Solution())->minCut(str) << endl;return 0;
}

　　

I

第一题用动态规划也是可以做的, 不过会比较麻烦(与Word Break类似)

这里用 dfs 加打印路径, 比较直观

int palindrom[1500][1500];
vector<vector<string> > res;
class Solution {
public:vector<vector<string>> partition(string s) {res.clear();memset(palindrom, 0x3f, sizeof(palindrom));vector<string> tmp;dfs(s, tmp, 0);return res;}bool isPalindrome(string &s, int i, int j) {if(palindrom[i][j] == 1)return true;if(j <= i)return (palindrom[i][j] = true);if(palindrom[i][j] == 0)return false;return (palindrom[i][j] = (s[i]==s[j] && isPalindrome(s,i+1,j-1)));}void dfs(string &s, vector<string> cur_vec, int depth) {if(depth == s.size()) {res.push_back(cur_vec);return;}for(int i = depth; i < s.size(); i ++) {if(isPalindrome(s, depth,i)) {cur_vec.push_back(s.substr(depth,i-depth+1));dfs(s, cur_vec, i+1);cur_vec.pop_back();}}}
};