leetcode - Palindrome Partitioning II-优快云博客

本文链接：https://blog.youkuaiyun.com/guixunlong/article/details/8825316

本文介绍了一种使用动态规划解决最小回文切分问题的方法，通过递归与记忆化搜索来减少重复计算，并最终采用递推法实现了高效求解。

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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.

动态规划：

记忆化搜索实现，能pass small data set，但运行large data set里 TLE了。

public class Solution {
    private java.util.Map<String, Integer> map = new java.util.HashMap<String, Integer>();
    private java.util.Map<String, Boolean> palinMap = new java.util.HashMap<String, Boolean>();
    public int minCut(String s) {
       if (s.length() == 1)
            return 0;
        
        if (map.get(s) != null)
            return map.get(s);
        
        if (isPalindrome(s)) {
            map.put(s, 0);
            return 0;
        }
        
        int ret = s.length()-1;
        
        for (int i=1; i<s.length(); i++) {
            ret = Math.min(ret, minCut(s.substring(0, i))+1+minCut(s.substring(i)));
        }
        map.put(s, ret);
        return ret;
    }
    
    public boolean isPalindrome(String s) {
        if (palinMap.get(s) != null)
            return palinMap.get(s);
        boolean ret = true;
        int i=0, j=s.length()-1;
        while (i < j) {
            char b = s.charAt(i++);
            char e = s.charAt(j--);
            if (b != e) {
                ret = false;
                break;
            }
        }
        palinMap.put(s, ret);
        return ret;
    }
}

这里采用了记忆化搜索方法记录了字符串是否为回文串的信息，但依然TLE，看样子得想别的办法了，该算法改用递推法实现为：

public class Solution {
    
    private java.util.Map<String, Boolean> palinMap = new java.util.HashMap<String, Boolean>();
    public int minCut(String s) {
       int len = s.length();
       int[][] data = new int[len][len];
       for (int i=0; i<len; i++)
            for (int j=0; j<len; j++)
                if (i==j)
                    data[i][j] = 0;
                else
                    data[i][j] = s.length()-1;
       for (int add=1; add<len; add++)
            for (int i=0; i<len&&(i+add)<len; i++) {
                 if (isPalindrome(s.substring(i, i+add+1))) {
                     data[i][i+add] = 0;
                 } else {
                     for (int d=1; d<=add; d++) {
                         data[i][i+add] = Math.min(data[i][i+add], data[i][i+d-1]+data[i+d][i+add]+1);
                    }
                 }
            }
        return data[0][len-1];
    }
    
    public boolean isPalindrome(String s) {
        if (palinMap.get(s) != null)
            return palinMap.get(s);
        boolean ret = true;
        int i=0, j=s.length()-1;
        while (i < j) {
            char b = s.charAt(i++);
            char e = s.charAt(j--);
            if (b != e) {
                ret = false;
                break;
            }
        }
        palinMap.put(s, ret);
        return ret;
    }
}