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();
            }
        }
    }
};