算法--地点线路问题
算法--地铁线路问题
废话不说，上代码，都有注释，就不解释什么了
import java.util.Scanner; public class train { public static String[] one = {"A10", "A11", "A12", "A13", "A14", "A15", "A16", "A17", "A18", "A19", "A20", "A21", "A22", "A23", "A24", "A25", "A26"}; public static String[] two = {"B1", "B2", "B3", "A12", "B4", "B5", "B6", "B7", "B8","B9", "B10", "A24", "B11", "B12", "B13", "B14"}; public static void main(String[] args) { Scanner sc = new Scanner(System.in); System.out.println("请输入开始站"); String trainbegin = sc.next(); System.out.println("请输入结束站"); String trainend = sc.next(); train_fround_minload(trainbegin,trainend); } /** * 判断属于哪个类型，进行调用 * 1.全部为A开头 * 2.有A有B开头 * 3.全部为B开头 * 注：将AB和BB开头分开，可以增加代码可读性 * @param trainbegin * @param trainend */ public static void train_fround_minload(String trainbegin,String trainend){ switch (analyse(trainbegin,trainend)){ case "AA":AA(trainbegin,trainend); ;break; case "AB":AB(trainbegin,trainend);break; case "BB":BB(trainbegin,trainend); ;break; default:break; } } /** * * @param trainbegin 开始站次 * @param trainend 结束站次 */ public static void AA(String trainbegin,String trainend){ String[] as1 = trainbegin.split("A"); String[] as2 = trainend.split("A"); int be =Integer.parseInt(as1[1]); int en = Integer.parseInt(as2[1]); if (be < 10 || en > 26){ System.out.println("err:请输入航班线内的站次!"); }else { int len = en - be; System.out.println(); System.out.println("------------------"); System.out.println("一共走了" +Math.abs(len) + "站"); System.out.println("分别是:"); if (be == en ){ System.out.println("请下车"); }else{ if (en > be){ for (int i = 0; i <= len; i++) { System.out.print("A" + (be + i) + " "); } }else{ for (int i = 0; i <= Math.abs(len); i++) { System.out.print("A" + (be - i) + " "); } } } } } public static void AB(String trainbegin,String trainend){ //如果是A12或者A24的，就说明是BB的情况，调用BB if (trainbegin.equals("A12") || trainbegin .equals( "A24") || trainend .equals("A12") ||trainend .equals( "A24")){ BB(trainbegin,trainend); }else{ //首先运用拆分看第一个是否为A，因为其他情况都已经讨论过了，所以，只有AB或BA的两种情况 int betwo = printArray(two, trainbegin); int entwo = printArray(two, trainend); String[] as1 = trainbegin.split("A"); //情况--AB 1.as[1].length 的长度等于2代表是A开头，反之不是 if (as1[1].length() == 2){ int beone = printArray(one, trainbegin); if (beone <= 2 ){ AA(trainbegin,"A12"); AB("A12",trainend); }else{ if (beone >= 14){ AA(trainbegin,"A24"); AB("A24",trainend); }else{ if ((beone-2)+minlen("A12",trainend) <= (20- ((beone-2)+minlen("A12",trainend)))){ AA(trainbegin,"A12"); AB("A12",trainend); }else{ AA(trainbegin,"A24"); AB("A24",trainend); } } } }else{ int enone = printArray(one, trainend); if (enone <= 2){ AB(trainbegin,"A12"); AA("A12",trainend); }else{ if(enone >= 14){ AB(trainbegin,"A24"); BB("A24",trainend); }else{ if (minlen("A12",trainbegin)+(enone-2) <= (20 - (minlen("A12",trainbegin)+(enone-2)))){ BB(trainbegin,"A12"); AB("A12", trainend); }else{ BB(trainbegin,"A24"); AB("A24",trainend); } } } } } } public static int minlen(String what,String trainplace){ int i = -1; if(what.equals("A12")){ i = printArray(two,trainplace) - printArray(two,what); if (i>0){ i = Math.min(i,16-i); return i; }else{ i = printArray(two,what) - printArray(two,trainplace); i = Math.min(i,16-i); return i; } } System.out.println("error!"); return i; } /** * 全部以B开头的一定是走二号线，所以单独列出来，进行逻辑的判断，这个也可以有AB开头的进行调用 * @param trainbegin * @param trainend */ public static void BB(String trainbegin,String trainend){ int be = printArray(two,trainbegin); int en = printArray(two,trainend); //将问题拆成下标大于8和小于8，分别对应的是按数组走和不按数组走 if (Math.abs(en-be) <= 8 ){ //如果en大于be就代表正向，小于代表逆向 System.out.println("一共走了"+Math.abs(en-be)+"站"); if (en > be){ System.out.println("分别是:"); for (int i = 0 ;i <= Math.abs(en-be);i++){ System.out.print(two[be+i]+" "); } }else{ System.out.println("分别是:"); for (int i = 0 ;i <= Math.abs(en-be);i++){ System.out.print(two[be-i]+" "); } } }else{ int count = Math.abs(Math.abs(en - be)-16); System.out.println("一共走了"+count+"站"); if (en > be){ for (int i = 0 ;i <= count ; i++){ if (be - i >= 0) { System.out.print(two[be - i]+" "); } else{ System.out.print(two[16+be-i]+" "); } } }else{ for (int i = 0 ;i <= count; i++){ if (be + i <= 15){ System.out.print(two[be+i]+" "); }else{ System.out.print(two[be+i-16]+" "); } } } } } /** * * @param trainbegin * @param trainend * @return 判断相对应的类型，为找到返回null */ public static String analyse(String trainbegin,String trainend) { String sbegin = String.valueOf(trainbegin.charAt(0)); String send = String.valueOf(trainend.charAt(0)); if (sbegin.equals("A") && send .equals("A")) { return "AA"; } else if ((sbegin .equals("A") && send .equals("B")) || (sbegin .equals("B") && send .equals("A"))) { return "AB"; } else if (sbegin .equals("B") && send.equals("B")) { return "BB"; } return null; } /** * * @param array 要查找的数组 * @param value 查找数组中的元素 * @return 找到返回下标，未找到返回-1 */ public static int printArray(String[] array,String value){ for(int i = 0;i<array.length;i++){ if(array[i].equals(value)){ return i; } } return -1;//当if条件不成立时，默认返回一个负数值-1 } }
这个算法的更优解是使用广度优先算法，创建一个二维数组，里面分别放入相应的左右连接，然后依此查找，会比我写的直接查找方便很多，这个简单问题可以剩下来50行代码左右吧，越复杂，就越简洁
