广联达2016西安研发笔试题

1. 华为机试考过的字符串压缩问题，没有难度。
2. 给出了平衡二叉树的概念，让写一个函数判断是否为平衡二叉树。剑指offer书本的原题
   方法一：

struct BinaryTreeNode{    int m_Value;    BinaryTreeNode* m_pLeft;    BinaryTreeNode* m_pRight;};int TreeDepth(BinaryTreeNode* pRoot){    if (pRoot == NULL)        return 0;    int nLeftDepth = TreeDepth(pRoot->m_pLeft);    int nRightDepth = TreeDepth(pRoot->m_pRight);    return (nLeftDepth>nRightDepth)?(nLeftDepth+1):(nRightDepth+1);}bool IsBalanced(BinaryTreeNode* pRoot){    if(pRoot== NULL)        return true;    int nLeftDepth = TreeDepth(pRoot->m_pLeft);    int nRightDepth = TreeDepth(pRoot->m_pRight);    int diff = nRightDepth-nLeftDepth;    if (diff>1 || diff<-1)        return false;    return IsBalanced(pRoot->m_pLeft)&&IsBalanced(pRoot->m_pRight);}

方法二：

bool IsBalanced(BinaryTreeNode* pRoot, int* depth){    if(pRoot== NULL)    {        *depth = 0;        return true;    }    int nLeftDepth,nRightDepth;    bool bLeft= IsBalanced(pRoot->m_pLeft, &nLeftDepth);    bool bRight = IsBalanced(pRoot->m_pRight, &nRightDepth);    if (bLeft && bRight)    {        int diff = nRightDepth-nLeftDepth;        if (diff<=1 || diff>=-1)        {            *depth = 1+(nLeftDepth > nRightDepth ? nLeftDepth : nRightDepth);            return true;        }    }    return false;}bool IsBalanced(BinaryTreeNode* pRoot){    int depth = 0;    return IsBalanced(pRoot, &depth);}

1. 找到两个单链表的交点，如果不想交返回空，还有判断链表是否有环都是链表最基本的操作，剑指offer都有非常类似的题。
   参考：http://blog.chinaunix.net/uid-20754793-id-177773.html
2. 跟广联达业务比较紧密的一道题，其实就是建立业务描述和真实数据映射关系，题目比较长思路简单。
3. 输入一个矩阵，顺时针打印矩阵的数据。也是剑指offer原题

package junit;public class PrintMatrix {    public static void printCircle(int[][] matrix, int startX, int startY, int endX, int endY) {        // only one column left        if (startY == endY) {            for (int i = startX; i <= endX; i++ ) {                System.out.println(matrix[i][endY]);            }            return;        }        // only one row left        if (startX == endX) {            for (int i = startY; i <= endY; i++ ) {                System.out.println(matrix[startX][i]);            }            return;        }        // 注意交界点不要打印两次        for (int i = startY; i < endY; i++ ) {            System.out.println(matrix[startX][i]);        }        for (int i = startX; i < endX; i++ ) {            System.out.println(matrix[i][endY]);        }        for (int i = endY; i > startY; i-- ) {            System.out.println(matrix[endX][i]);        }        for (int i = endX; i > startX; i-- ) {            System.out.println(matrix[i][startY]);        }    }    public static void printMatrix(int[][] matrix) {        if (matrix == null) {            return;        }        int startX = 0;        int startY = 0;        int endY = matrix[0].length - 1;        int endX = matrix.length - 1;        while ((startX <= endX) && (startY <= endY)) {            printCircle(matrix, startX, startY, endX, endY);            startX++;            startY++;            endX--;            endY--;        }    }    public static void main(String[] args) {        int[][] matrix = {{1,2,3,4},{5,6,7,8},{9,10,11,12},{13,14,15,16}};        printMatrix(matrix);    }}

1. 两个版本号比较，就是字符串的一些处理，java 使用split 函数切分比较就轻松完成。这里版本号比较只有数字和. 所以难度比字母存在更低。

package junit;public class CompareVersion {    public static int compare_string(String str1, String str2) {        // 特殊字符一定要注意转义，是个坑        String[] strs1 = str1.split("\\.");        String[] strs2 = str2.split("\\.");        int i, j;        int num1 ,num2;        for (i = 0, j = 0; i < strs1.length && j < strs2.length; i++, j++) {            num1 = Integer.parseInt(strs1[i]);            num2 = Integer.parseInt(strs2[j]);            if(num1 > num2)                return 1;            if(num1 < num2)                return -1;        }        if (i == strs1.length && j == strs2.length) {            return 0;        }        if (i == strs1.length) {            return -1;        } else {            return 1;        }    }    public static void main(String[] args) {        String version1 = "1.0.1";          String version2 = "1.0";        System.out.println(compare_string(version1, version2));    }}

其他一些题目也基本都是来自剑指offer。