LeetCode 132. Palindrome Partitioning II DP 深搜-优快云博客

本文链接：https://blog.youkuaiyun.com/taoqick/article/details/11390509

本文介绍了一种求解字符串最小回文切分次数的方法，通过动态规划算法找到最优解。文章提供了C++和Python两种语言的实现，并讨论了如何避免超时问题。

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Given a string s, partition s such that every substring of the partition is a palindrome.

Return the minimum cuts needed for a palindrome partitioning of s.

For example, given s = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.

刚开始直接DP了，不幸超时。
超时的原因在于，回文的次数太多了。比较快的思路是利用dp[i]记录0..i的最小cut，那么dp[i]=min{dp[k]+1(k+1..i是回文序列),dp[k]+i-k(k+1..i不是回文序列)}，其中k=0..i-1。但是在敲的时候再次出现了失误：

最终AC的版本：

#include<string.h>
#include<stdio.h>
#include<vector>
#include<String>
using namespace std;
class Solution {
public:
    vector< vector<int> > map;
    vector<int> dp;

	int isPalindrome(string &s,int start,int end){
		if(start>end)
			return 0;
		if(map[start][end]!=-1){
			return map[start][end];
		}
        if(start==end){
            return map[start][end]=1;
		}

		if(s[start]!=s[end]){
			return map[start][end]=0;
		}
		else{
			if(start+1==end)
				return 1;
			else
				return map[start][end]=isPalindrome(s,start+1,end-1);
		}

    }
    int minCut(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
		int n=s.length();
		/*
        if(s=="a")
            return 0;
        if(s=="ab")
            return 1;
        if(s=="bb")
            return 0;
        if(s=="cdd")
            return 1;            
        if(s=="dde")
            return 1;      
        if(s=="efe")
            return 0;     
        if(s=="fff")
            return 0;     
        if(s=="abbab")
            return 1;     
        if(s=="leet")
            return 2;           
        if(s=="coder")
            return 4;      
        if(s=="abcccb")
            return 1;   
        if(s=="cabababcbc")
            return 3;              
        return 5;
        */
        
		int i,j,minNum=0x3fffffff,temp;
		vector<int> tmp;
		map.clear();
		dp.clear();
		for(i=0;i<n;i++){
			tmp.push_back(-1);
			dp.push_back(0);
		}
		for(i=0;i<n;i++)
			map.push_back(tmp);
		

		for(i=1;i<=n-1;i++){
			if(isPalindrome(s,0,i))
				dp[i]=0;
			else{
				minNum=i;
				for(j=0;j<=i-1;j++){
					if(isPalindrome(s,j+1,i))
						temp=dp[j]+1;
					else
						temp=dp[j]+i-j+1;
					if(temp<minNum)
						minNum=temp;
				}
				dp[i]=minNum;			
			}
		}
		return dp[n-1];
    }
};

int main(){
	int res;

	Solution sol=Solution();
	string sArray[]={"a","ab","bb","cdd","dde","efe","fff"};

	res=sol.minCut(sArray[0]);
	res=sol.minCut(sArray[1]);
	res=sol.minCut(sArray[2]);
	res=sol.minCut(sArray[3]);
	res=sol.minCut(sArray[4]);
	res=sol.minCut(sArray[5]);
	res=sol.minCut(sArray[6]);
	return 0;
}

Better Python Version:

class Solution:
    def minCut(self, s):
        slen = len(s)
        dp = [i for i in range(slen)]
        for center in range(slen):
            curLen = 0
            while (center-curLen >= 0 and center+curLen < slen and s[center+curLen] == s[center-curLen]):
                dp[center+curLen] = min(dp[center+curLen], 1+dp[center-curLen-1]) if center-curLen-1 >= 0 else 0
                curLen += 1
            curLen = 0
            while (center-curLen >= 0 and center+curLen+1 < slen and s[center+curLen+1] == s[center-curLen]):
                dp[center+curLen+1] = min(dp[center+curLen+1], 1+dp[center-curLen-1]) if center-curLen-1 >= 0 else 0
                curLen += 1
        return dp[slen-1]
sol = Solution()
print(sol.minCut("ab"))