LeetCode 514----Freedom Trail

问题描述

In the video game Fallout 4, the quest "Road to Freedom" requires players to reach a metal dial called the "Freedom Trail Ring", and use the dial to spell a specific keyword in order to open the door.

Given a string ring, which represents the code engraved on the outer ring and another string key, which represents the keyword needs to be spelled. You need to find the minimum number of steps in order to spell all the characters in the keyword.

Initially, the first character of the ring is aligned at 12:00 direction. You need to spell all the characters in the string key one by one by rotating the ring clockwise or anticlockwise to make each character of the string key aligned at 12:00 direction and then by pressing the center button.

At the stage of rotating the ring to spell the key character key[i]:

1. You can rotate the ring clockwise or anticlockwise one place, which counts as 1 step. The final purpose of the rotation is to align one of the string ring's characters at the 12:00 direction, where this character must equal to the character key[i].
2. If the character key[i] has been aligned at the 12:00 direction, you need to press the center button to spell, which also counts as 1 step. After the pressing, you could begin to spell the next character in the key (next stage), otherwise, you've finished all the spelling.

Example:

Input: ring = "godding", key = "gd"
Output: 4
Explanation:
 For the first key character 'g', since it is already in place, we just need 1 step to spell this character. 
 For the second key character 'd', we need to rotate the ring "godding" anticlockwise by two steps to make it become "ddinggo".
 Also, we need 1 more step for spelling.
 So the final output is 4.

Note:

1. Length of both ring and key will be in range 1 to 100.
2. There are only lowercase letters in both strings and might be some duplcate characters in both strings.
3. It's guaranteed that string key could always be spelled by rotating the string ring.

算法分析：

该题需要使用动态规划算法进行计算  
先声明一个二维的动态规划表 int [][] dp=new int[key.length()][ring.length()]; dp[i][j] 的值表示转动到 key 中的第 i 个字符在 ring 中第 j 个位置时总共需要的转动 step （不包括按下 button 键的 step )。  
状态转移方程：  
dp[i][j]=min(dp[i][j],dp[i-1][k]+min(abs(j-k),ring.leng()-abs(j-k))).
该方程中，i 表示当前考察的 key 中的字符在 key 中的下标，j 表示当前考察的字符在 ring 中的下标，k 表示上一个考察的字符在 ring 中的下标。

Java 算法实现：

public class Solution {
    public int findRotateSteps(String ring, String key) {
        ArrayList<Integer>[] list=new ArrayList[128];
		char ch;
		for(int i=0;i<ring.length();i++){//统计ring中所有的字符在ring中的下标
			ch=ring.charAt(i);
			if(list[ch]==null){
				list[ch]=new ArrayList<Integer>();
			}
			list[ch].add(i);
		}

		int ringLoop=ring.length();
		int[][]dp=new int[key.length()][ring.length()];
		for(Integer index:list[key.charAt(0)]){//对 key 中第一个字符在 ring 中的位置进行记录
			dp[0][index]=Math.min(index, ringLoop-index);
		}
		char former=key.charAt(0),cur;
		for(int i=1;i<key.length();i++){//动态规划算法
			cur=key.charAt(i);
			for(Integer indexCur:list[cur]){//遍历当前考察的字符在 ring 中的位置
				dp[i][indexCur]=Integer.MAX_VALUE;
				for(Integer indexFormer:list[former]){//遍历上一个字符在 ring 中的位置
					int distance=Math.abs(indexCur-indexFormer);//当前考察的字符与上一个考察的字符在 ring 中的距离
					dp[i][indexCur]=Math.min(dp[i][indexCur], dp[i-1][indexFormer]+Math.min(distance,ringLoop-distance));
				}
			}
			former=cur;
		}
		int mindist=Integer.MAX_VALUE;
		int depth=key.length()-1;// key 中最后一个字符的下标
		for(Integer lastIndex:list[key.charAt(depth)]){//遍历 key 中最后一个字符在 ring 中的所有位置
			if(mindist>dp[depth][lastIndex]){ //找到到达 key 中最后一个字符所需的最小 step 数
				mindist=dp[depth][lastIndex];
			}
		}
		return mindist+key.length(); //总的 step 数要包括按下 button 键的次数
    }
}