ac

给定一个字符串数组strs，再给定两个字符串str1和str2，返回在strs中str1和str2的最小距离，如果str1或str2为null，或不在strs中，返回-1。

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    public static Scanner sc = new Scanner(System.in);

    public static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

    public static void main(String[] args) throws Exception {
        String nStr = reader.readLine();
        int n = Integer.valueOf(nStr);
        String s12 = reader.readLine();
        String[] sArr = s12.split(" ");
        String[] strs = new String[n];
        for(int i=0;i<n;i++){
            strs[i] = reader.readLine();
        }
        if(n <= 1 || sArr.length <= 1){
            System.out.println(-1);
            return;
        }
        String s1 = sArr[0];
        String s2 = sArr[1];

        int lastS1Pos = -1;
        int lastS2Pos = -1;
        int ans = -1;

        for(int i=0;i<n;i++){
            if(strs[i].equals(s1)){
                if(lastS2Pos != -1){
                    // 计算完并清除s2
                    if(ans == -1) {
                        ans = i - lastS2Pos;
                    }else {
                        ans = Math.min(ans, i - lastS2Pos);
                    }
                    lastS2Pos = -1;
                    lastS1Pos = i;
                }else {
                    lastS1Pos = i;
                }
            }
            if(strs[i].equals(s2)){
                if(lastS1Pos != -1){
                    // 计算完并清除 s1
                    if(ans == -1) {
                        ans = i - lastS1Pos;
                    }else {
                        ans = Math.min(ans, i - lastS1Pos);
                    }
                    lastS1Pos = -1;
                    lastS2Pos = i;
                }else {
                    lastS2Pos = i;
                }
            }
        }
        System.out.println(ans);
    }

}
/*
1
CD AB
CD

-1


5
QWER 666
QWER
1234
qwe
666
QWER

1

5
QWER kkkk
QWER
1234
qwe
666
QWER
*/