Palindrome Partitioning II

最新推荐文章于 2025-12-02 15:55:42 发布

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本文介绍了一种使用动态规划解决字符串最小切割问题的方法，目的是将字符串分割成多个子串，确保每个子串都是回文串，并尽可能减少切割次数。

大例子还是没通过，先放着

public class Solution {
    int[][] is_sub_palind;
	
	public int minCut(String s){
		int n = s.length();
		int[] p = new int[n+1];
		is_sub_palind= new int[n+1][n+1];
		for(int i=0;i<n+1;i++){
			 p[i] = 1000000;
            for(int j=0;j<n+1;j++){         
                is_sub_palind[i][j] = -1;
            }
		}
		p[0]=0;
		for(int i=0;i<=n;i++){
			int low = 10000000;
			for(int k=0;k<i;k++){
				if(isPalindrome(s,k+1,i)){
					if(k==0)p[i]=0;
					else{p[i]=p[k]+1;}
				}
				else{
					p[i]=p[k]+i-k;
				}
				if(p[i]<low){
					low = p[i];
				}		
			}
			p[i] = low;
			//System.out.println(p[i]);
		}
		return p[n];
		
	}
	
	
	/*	
    public int minCut(String s) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int n = s.length();
       
        int[][] p = new int[n+1][n+1];
        is_sub_palind= new int[n+1][n+1];
       
        for(int i=0;i<n+1;i++)
            for(int j=0;j<n+1;j++){
                p[i][j] = 1000000;
                is_sub_palind[i][j] = -1;
            }
        //if(isPalindrome(s,1,n)==true)return 0;
        for(int k=0;k<=n-1;k++){
            for(int i=1;i<=n-k;i++){
            	int sum=p[i][i+k];
                //if(is_sub_palind[i][i+k]==1){
            	if(isPalindrome(s,i,i+k)==true){
                	p[i][i+k]=0;
                }
                else{
                	for(int t=0;t<k;t++){
                		sum = p[i][i+t]+p[i+t+1][i+k]+1;
                		if(sum<p[i][i+k])
                			p[i][i+k]=sum;
                	}
                }
            }
        }
       
        return p[1][n];
    }
    */
    private boolean isPalindrome(String s,int i,int j){
    	if(i==j)return true;
    	if(i>j)return false;
    	if(is_sub_palind[i][j]==1)return true;
    	if(is_sub_palind[i][j]==0)return false;
    	int i1=i-1;
    	int j1=j-1;
    	while(i1<j1){
    		if(s.charAt(i1++)==s.charAt(j1--))continue;
    		is_sub_palind[i][j]=0;
    		return false;
    	}
    	is_sub_palind[i][j]=1;
    	return true;
    }
    
    public static void main(String[] args){
    	//String s= "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa";
    	 String s = "dde";
    	System.out.println(s.length());
    	Solution a = new Solution();
    	int t = a.minCut(s);
    	System.out.println(t);
    }
}

public class Solution {
	int[][] is_sub_palind;
	
	public int minCut(String s){
		int n = s.length();
		int[] p = new int[n+1];
		is_sub_palind= new int[n+1][n+1];
		for(int i=0;i<n+1;i++){
			 p[i] = i;
            for(int j=0;j<n+1;j++){         
                is_sub_palind[i][j] = -1;
            }
		}
		for(int i=1;i<=n;i++){
			int low = p[i];
			for(int k=0;k<i;k++){
				if(isPalindrome(s,k+1,i)){
					if(k==0)p[i]=0;
					else{p[i]=p[k]+1;}
				}
				else{
					p[i]=p[k]+i-k;
				}
				if(p[i]<low){
					low = p[i];
				}	
			}
			p[i] = low;	
		}
		return p[n];
	}

    private boolean isPalindrome(String s,int i,int j){
  
    	if(i==j)return true;
    	if(i>j)return false;
    	
    	if(is_sub_palind[i][j]==1)return true;
    	
    	if(is_sub_palind[i][j]==0){if(i==1&&j==2)return false;}
    	
    	if(s.charAt(i-1)==s.charAt(j-1)){
    		if(i+1<j-1){
    			if(is_sub_palind[i+1][j-1]==1){is_sub_palind[i][j]=1;return true;}
    			else {is_sub_palind[i][j]=0;return false;}
    		}
    		else {is_sub_palind[i][j]=1;return true;}
    	}
    	is_sub_palind[i][j]=0;
    	return false;
    }
    
    public static void main(String[] args){
    	//String s= "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa";
    	 String s = "aabaae";
    	//System.out.println(s.length());
    	Solution a = new Solution();
    	long startTime=System.currentTimeMillis();
    	int t = a.minCut(s);
    	long endTime=System.currentTimeMillis(); 
    	
    	System.out.println(t);
    	System.out.println("程序运行时间： "+(endTime-startTime)+"ms");
    }
}

查找是否是回文串也要用dp。