【LeetCode】664.Strange Printer（hard）解题报告

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本文介绍了解决LeetCode上Strange Printer问题的一种动态规划方法。该问题要求找出打印给定字符串所需的最少轮数，打印机只能逐个字符打印相同字符，并且可以覆盖原有字符。通过构建二维DP表格并利用子问题最优性来降低时间复杂度。

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【LeetCode】664.Strange Printer(hard)解题报告

tags: DP

题目地址：https://leetcode.com/problems/strange-printer/description/
题目描述：

There is a strange printer with the following two special requirements:
1. The printer can only print a sequence of the same character each time.
2. At each turn, the printer can print new characters starting from and ending at any places, and will cover the original existing characters.
Given a string consists of lower English letters only, your job is to count the minimum number of turns the printer needed in order to print it.

Example1：

Input: "aaabbb"
Output: 2
Explanation: Print "aaa" first and then print "bbb".

Example2:

Input: "aba"
Output: 2
Explanation: Print "aaa" first and then print "b" from the second place of the string, which will cover the existing character 'a'.

分析：

s:       abc......d
index:   i    k   j
f(i,j): min steps to print s
f(i,j): worse case j-i+1
f(i,j): min(f(i,k) + f(k+1,j) - 1 if(s[k] == s[j] {i <= k < j}))
aba

Solutions:

class Solution {
    public int strangePrinter(String s) {
        if(s.length()==0){
            return 0;
        }
        int len = s.length();
        int[][] dp = new int[len][len];
        for(int i=0 ; i<len ; i++){
            dp[i][i] = 1 ;
        }
        for(int j=1 ; j<len ; j++){
            for(int i=0 ; i+j<len ; i++){
                dp[i][i+j] = j+1 ;
                for(int k=i ; k<i+j ; k++){
                    int total = dp[i][k] + dp[k+1][i+j];
                    if(s.charAt(k) == s.charAt(i+j)){
                        total--;
                    }
                    dp[i][i+j] = Math.min(dp[i][i+j],total);
                }
            }
        }
        return dp[0][len-1];
    }
}

Date：2017年11月5日