Leetcode: Palindrome Partition I II
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();
}
}
}
};