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2014
11-18

LeetCode-Palindrome Partitioning II[动态规划]

Palindrome Partitioning II

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.

标签: Dynamic Programming
分析

定义状态{f(i,j)}表示区间{[i,j]}之间最小的cut数,则状态转移方程为
$$
f(i,j)=\min\left\{f(i,k)+f(k+1,j)\right\}, i \leq k \leq j, 0 \leq i \leq j<n
$$
这是一个二维函数,实际写代码比较麻烦。

所以要转换成一维DP。如果每次,从i往右扫描,每找到一个回文就算一次DP的话,就可以转换为{f(i)=区间[i, n-1]之间最小的cut数},n为字符串长度,则状态转移方程为
$$
f(i)=\min\left\{f(j+1)+1\right\}, i \leq j<n
$$

一个问题出现了,就是如何判断{[i,j]}是否是回文?每次都从i到j比较一遍?太浪费了,这里也是一个DP问题。

定义状态{P[i][j] = true if [i,j]为回文},那么
\begin{Code}
P[i][j] = str[i] == str[j] && P[i+1][j-1]
\end{Code}

代码1

// LeetCode, Palindrome Partitioning II
// 时间复杂度O(n^2),空间复杂度O(n^2)
class Solution {
public:
    int minCut(string s) {
        const int n = s.size();
        int f[n+1];
        bool p[n][n];
        fill_n(&p[0][0], n * n, false);
        //the worst case is cutting by each char
        for (int i = 0; i <= n; i++)
            f[i] = n - 1 - i; // 最后一个f[n]=-1
        for (int i = n - 1; i >= 0; i--) {
            for (int j = i; j < n; j++) {
                if (s[i] == s[j] && (j - i < 2 || p[i + 1][j - 1])) {
                    p[i][j] = true;
                    f[i] = min(f[i], f[j + 1] + 1);
                }
            }
        }
        return f[0];
    }
};

Java代码:

public class Solution {
    public int minCut(String s) {
			int n = s.length();
			boolean isPal[][] = new boolean[n][n];
			int dpminc[] = new int[n];
			for (int j = 0; j < n; j++){
				dpminc[j] = j;
				for (int i = 0; i <= j; i++){
					if(s.charAt(i) == s.charAt(j) && (j - i <= 1 || isPal[i+1][j-1]) ){
						isPal[i][j] = true;
						if(i > 0)
							dpminc[j] = Math.min(dpminc[j], dpminc[i - 1] + 1);
						else
							dpminc[j] = 0;
					}
				}
			}
			return dpminc[n-1];	
		}
}

相关题目
Palindrome Partitioning


  1. #include <cstdio>
    #include <cstring>

    const int MAXSIZE=256;
    //char store[MAXSIZE];
    char str1[MAXSIZE];
    /*
    void init(char *store) {
    int i;
    store['A']=’V', store['B']=’W',store['C']=’X',store['D']=’Y',store['E']=’Z';
    for(i=’F';i<=’Z';++i) store =i-5;
    }
    */
    int main() {
    //freopen("input.txt","r",stdin);
    //init(store);
    char *p;
    while(fgets(str1,MAXSIZE,stdin) && strcmp(str1,"STARTn")==0) {
    if(p=fgets(str1,MAXSIZE,stdin)) {
    for(;*p;++p) {
    //*p=store[*p]
    if(*p<’A’ || *p>’Z') continue;
    if(*p>’E') *p=*p-5;
    else *p=*p+21;
    }
    printf("%s",str1);
    }
    fgets(str1,MAXSIZE,stdin);
    }
    return 0;
    }

  2. 嗯 分析得很到位,确实用模板编程能让面试官对你的印象更好。在设置辅助栈的时候可以这样:push时,比较要push的elem和辅助栈的栈顶,elem<=min.top(),则min.push(elem).否则只要push(elem)就好。在pop的时候,比较stack.top()与min.top(),if(stack.top()<=min.top()),则{stack.pop();min.pop();},否则{stack.pop();}.