[leetcode] Palindrome Partitioning II

最新推荐文章于 2019-06-01 11:25:07 发布

iteye_15479

最新推荐文章于 2019-06-01 11:25:07 发布

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分类专栏： algorithm java 文章标签：开发工具

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本文介绍了一个动态规划问题：如何计算将字符串分割成回文子串所需的最少切割次数，并提供了一段Java实现代码。

http://leetcode.com/onlinejudge#question_132

Question:

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.

Solution:

DP problem, but my first try failed due to time limitation. The root cause is you should try to optimize the Palindrome calculation. The solution below borrows from others idea found from the discussion page. Thanks to the author.


public class Solution {
    public int minCut(String s) {
        int leng = s.length();
        if(leng == 0 || leng == 1) return 0;
        boolean[][] isPal = new boolean[leng][leng];

        int[] dp = new int[leng]; 
        for (int i = 0; i < leng; i++) {
            dp[i] = leng - 1 - i;
        }

        for (int i = leng - 1; i >= 0; --i) {
            for (int j = i; j < leng; ++j) {
                if (s.charAt(i) == s.charAt(j) && (j <= i + 2 || (i + 1 < leng && j - 1 >= 0 && isPal[i + 1][j - 1]))) {
                    isPal[i][j] = true;
                    if(j+1 < leng){
                        dp[i] = Math.min(dp[i], 1 + dp[j + 1]);
                    }else {
                        dp[i] = 0;
                    }
                }
            }
        }

        return dp[0];
    }
    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Solution s = new Solution();
        int cut = s.minCut("abccdcba");
        System.out.println(cut);
    }

}