华为机试练习150817第二题：求后序序列

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import java.util.Scanner;/** * Created by cq on 2015/8/17 22:22. * 根据前序序列和中序序列，求后序序列，否则输出No * 这里没有对输入参数的合法性进行验证，因为合法性由机试系统保证 * 测试例： 8 1 2 4 7 3 5 6 8 4 7 2 1 5 3 8 6 8 1 2 4 7 3 5 6 8 4 1 2 7 5 3 8 6 8 1 2 4 5 3 6 7 8 4 5 2 1 6 3 9 7 * 结果： 7 4 2 5 8 6 3 1 No 5 4 2 6 9 7 3 1 */public class Main {    public static void main(String[] args){        Scanner in = new Scanner(System.in);        try {            while (in.hasNextInt()){                int lenOfTree = in.nextInt();                int[] preorder = new int[lenOfTree], inorder = new int[lenOfTree];                for (int i=0; i<lenOfTree; i++){                    preorder[i] = in.nextInt();                }                for (int i=0; i<lenOfTree; i++){                    inorder[i] = in.nextInt();                }                String result = isBTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);                if (result.indexOf("s") != -1){                    result = "No";                }                System.out.println(result);            }        }catch (Exception e){}        finally {            in.close();        }    }    public static String isBTree(int[] preorder, int pBegin, int pEnd, int[] inorder, int iBegin, int iEnd){        if (iBegin == iEnd){            return inorder[iBegin]+" ";        }        else if (iBegin > iEnd){            return "";        }        int root = preorder[pBegin], indexOfRoot = indexOf(inorder, iBegin, iEnd, root);        if (indexOfRoot == -1){            return "s";        }        int leftLen = indexOfRoot-iBegin;        return  isBTree(preorder, pBegin+1, pBegin+leftLen, inorder, iBegin, indexOfRoot-1)                + isBTree(preorder, pBegin+leftLen+1, pEnd, inorder, indexOfRoot+1, iEnd) +root+" ";    }    public static int indexOf(int[] arr, int froIndex, int endIndex, int target){        for (int i=froIndex; i<=endIndex; i++){            if (arr[i] == target){                return i;            }        }        return -1;    }}
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