palindrome-partitioning-ii

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题目描述

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”,
Return1since the palindrome partitioning[“aa”,”b”]could be produced using 1 cut.

class Solution {
public:
    bool palindrome(char *p,int len){
    char*q=p+len-1;

    while(q>p){
        if(*p==*q){
            q--;
            p++;
        }else{
            return false;
        }
    }

    return true;
}
    int minCut(string s) {
        int len=s.length(); 
    char *p=new char[s.length()+1];
    int ii;
    for( ii=0;ii<len;ii++){
        p[ii]=s[ii];
    }
    p[ii]='\0';


    //dp[i][j] 记录 从 index i 开始 长度为j的子串需要几次划分 
    int dp[s.length()][s.length()+1];

    for(int i=0;i<len;i++)
        dp[i][1]=0;

    //从第二列开始 
    for(int j=2;j<len+1;j++)
        for(int i=0;(i+j)<=len;i++){

            //如果加入新的元素仍是回文数 则仍只需0次划分 
            if(palindrome(p+i,j)){
                dp[i][j]=0;
            }else{
                //如果整个当前串不是回文 则从其划分中 找最小值
                int min=dp[i][j-1]+dp[i+j-1][1]+1;

                for(int k=j-2;k>=1;k--){
                    int temp=dp[i][k]+dp[i+k][j-k]+1;

                    min=min<temp?min:temp;

                } 
                dp[i][j]=min;
            }
        }
       return dp[0][len];
    }
};