题解 | #交错序列#

import java.util.Scanner;

public class Main {
    public static void main(String[] args){
        Scanner in = new Scanner(System.in);

        while(in.hasNext()){
            // solution1(in);
            // solution2(in);
            solution3(in);
        }
    }

    /**
     * 正则
     *
     * 连续多个0 等价于 一个0
     * 连续多个1 等价于 一个1
     *
     * @param in
     */
    private static void solution1(Scanner in){
        int n = Integer.parseInt(in.nextLine());
        // 去掉空格符
        String serial = in.nextLine().replaceAll(" ", "");

        // 多个0替换成一个0 多个1替换成一个1
        String result = serial.replaceAll("0{2,}", "0").replaceAll("1{2,}", "1");

        System.out.println(result.length());
    }

    /**
     * 动态规划
     *
     * dp[i]表示前i个字符能够得到的最长交错序列长度
     *
     *         { dp[i-1]  , char[i]=char[i-1]
     * dp[i] = {
     *         { dp[i-1]+1, char[i]!=char[i-1]
     *
     * @param in
     */
    private static void solution2(Scanner in){
        int n = Integer.parseInt(in.nextLine());
        // 去掉空格符
        String serial = in.nextLine().replaceAll(" ", "");

        int[] dp = new int[n];

        dp[0] = 1;
        for(int i=1; i<n; i++){
            if(serial.charAt(i) != serial.charAt(i-1)){
                dp[i] = dp[i-1]+1;
            }else{
                dp[i] = dp[i-1];
            }
        }

        System.out.println(dp[n-1]);
    }

    /**
     * 模拟法: 数组
     * @param in
     */
    private static void solution3(Scanner in){
        int n = in.nextInt();

        int[] nums = new int[n];

        int result = 0;
        for(int i=0; i<n; i++){
            nums[i] = in.nextInt();
            if(i == 0){
                result = 1;
            }else{
                if(nums[i] != nums[i-1]){
                    result++;
                }
            }
        }

        System.out.println(result);
    }
}