Strange Printer

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

题目：

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.

Example 1:

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

Example 2:

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'.

Hint: Length of the given string will not exceed 100.

解题思路：
这道题就是一个动态规划问题。
link： https://discuss.leetcode.com/topic/100240/java-o-n-3-dp-solution-with-explanation-and-simple-optimization

代码：

public int strangePrinter(String s) {
    if (s == null || s.length() == 0) {
        return 0;
    }

    int n = s.length();
    int[][] dp = new int[n][n];
    for (int i = 0; i < n; i++) {
        dp[i][i] = 1;
        if (i < n - 1) {
            dp[i][i + 1] = s.charAt(i) == s.charAt(i + 1) ? 1 : 2;
        }
    }

    for (int len = 2; len < n; len++) {
        for (int start = 0; start + len < n; start++) {
            dp[start][start + len] = len + 1;
            for (int k = 0; k < len; k++) {
                int temp = dp[start][start + k] + dp[start + k + 1][start + len];
                dp[start][start + len] = Math.min(
                        dp[start][start + len],
                        s.charAt(start + k) == s.charAt(start + len) ? temp - 1 : temp
                );
            }
        }
    }

    return dp[0][n - 1];
}