132. Palindrome Partitioning II-优快云博客

本文链接：https://blog.youkuaiyun.com/jimhust/article/details/60060947

本文介绍了一种算法，用于计算将字符串分割成回文子串所需的最少切割次数，并提供了两种解决方案，一种直接检查所有可能的回文子串，另一种通过预处理避免重复计算提高效率。

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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.

The idea is to maintain a result from i = 0 to (i - 1)

when the ith letter comes in dp[i] = min(dp[j - 1] + 1) for all the j that s(j -> i) is a palindrome .

Code:

public class Solution {
    public int minCut(String s) {
        int[] dp = new int[s.length() + 1];
        for(int i = 1; i < dp.length; i++){
            dp[i] = Integer.MAX_VALUE;
            for(int j = i; j > 0; j--){
                if(isPanlindrome(s,j - 1,i - 1)){
                    dp[i] = Math.min(dp[i],1 + dp[j - 1]);
                }
            }
        }
        return dp[dp.length - 1] - 1;
        
        
    }
    
    public boolean isPanlindrome(String s, int start, int end){
        while(start < end){
            if(s.charAt(start) == s.charAt(end)){
                start++;
                end--;
            } else {
                return false;
            }
        }
        return true;
    }
}

Time complexity O(N3). since

isPanlindrome() takes O(n)

This code will get TLE on Leetcode since for isPanlindrome part, there are many redundant computation.

For example "baaab" we already know that in the first "baaa" the aaa is a panlidrome. when the last b comes in , we should only check "b***b" and the info about "aaa" should be stored to improve the performance.

if s(i) != s(j) p[i][j] = false;

else p[i][j] = p[i - 1][j + 1].

code:

public class Solution {
    public int minCut(String s) {
        int[] dp = new int[s.length() + 1];
        boolean[][] p = new boolean[s.length()][s.length()];
        for(int i = 0 ; i < s.length(); i++){
            p[i][i] = true;
        }
        for(int i = 0 ; i < s.length() - 1; i++){
            p[i][i+1] = (s.charAt(i) == s.charAt(i+1));
        }
        
        for(int l = 2; l < s.length(); l++){
            for(int i = 0 ; i < s.length() - l; i++){
                int j = i + l;
                if(s.charAt(i) == s.charAt(j)){
                    p[i][j] = p[i + 1][j - 1];
                }
            }
        }
        
        
        for(int i = 1; i < dp.length; i++){
            dp[i] = Integer.MAX_VALUE;
            for(int j = i; j > 0; j--){
                if(p[j - 1][i - 1]){
                    dp[i] = Math.min(dp[i],1 + dp[j - 1]);
                }
            }
        }
        return dp[dp.length - 1] - 1;
        
        
    }

}

The pre-process of the data is important to avoid TLE, total Time O(n2).