state machine

本文主要是介绍state machine，对大家解决编程问题具有一定的参考价值，需要的程序猿们随着小编来一起学习吧！

You are given a string s consisting of lowercase letters and an integer k. We call a string t ideal if the following conditions are satisfied:

• t is a subsequence of the string s.
• The absolute difference in the alphabet order of every two adjacent letters in t is less than or equal to k.

Return the length of the longest ideal string.

A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

Note that the alphabet order is not cyclic. For example, the absolute difference in the alphabet order of 'a' and 'z' is 25, not 1.

 Example 1:

Input: s = "acfgbd", k = 2
Output: 4
Explanation: The longest ideal string is "acbd". The length of this string is 4, so 4 is returned.
Note that "acfgbd" is not ideal because 'c' and 'f' have a difference of 3 in alphabet order.

Example 2:

Input: s = "abcd", k = 3
Output: 4
Explanation: The longest ideal string is "abcd". The length of this string is 4, so 4 is returned.

 Constraints:

• 1 <= s.length <= 105
• 0 <= k <= 25
• s consists of lowercase English letters.

class Solution {
    int result = 1;
    public int longestIdealString(String s, int k) {
        //标示26个字母在前面出现的最后一个位置
        int[] memo = new int[s.length()];
        int[] pos = new int[26];
        int result = 0;
        //开始初始化为-1，因为所有字母都没出现过
        Arrays.fill(pos, -1);
        for(int i=0;i<s.length();i++){
            char c = s.charAt(i);
            //从  c-k ~ c+k ,逐个进行判定，看这些字母那个的最后一个位置最长
            for(int j = Math.max(0, c-'a'-k);j <= Math.min(25, c-'a'+k); j++){
                if(pos[j] != -1){
                    //在k范围内的最长字母长度基础上+1
                    memo[i] = Math.max(memo[i], memo[pos[j]]+1);
                }
                if(memo[i] == 0) memo[i] = 1;
            }
            result = Math.max(result, memo[i]);
            pos[c-'a'] = i;
        }
        return result;
    }
}

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