添加2011年准备面试和面试提解答代码

Author: wrchow

ms2011/Algorithm/lcs.cpp

@@ -0,0 +1,70 @@
+// dynamic plan for LCS problem 20111020 23:50~
+/* description:
+    1. construct a matrix of LCS path
+    2. find LCS in the path
+*/
+#include <iostream>
+#include <string>
+#include <vector>
+#include <stack>
+using namespace std;
+
+typedef vector<int> row;
+typedef vector<row> Matrix;
+
+void pathLCS(string A,string B,Matrix& c);
+void LCS(const Matrix& c);
+void printMatrix(const Matrix& c);
+
+int main(int argc,char *argv[]){
+    // find "ACD"  & "BCS"
+    string A = "ABCD";
+    string B = "BACD";
+    vector<int> t(A.size()+1,0);
+    Matrix c(B.size()+1,t);   
+    printMatrix(c);
+    pathLCS(A,B,c);
+    printMatrix(c);
+    LCS(c);
+    return 0;
+}
+
+void pathLCS(string A,string B,Matrix& c){
+    cout << "path matrix.." << endl;
+    int i,j;
+    for (i=1; i<=B.size(); i++) {
+        for (j=1; j<=A.size(); j++) {
+            if (B[B.size()-i+1] == A[A.size()-j+1]) // reverse A & B
+                c[i][j] = c[i-1][j-1] + 1;
+            else if(c[i-1][j] > c[i][j-1]) c[i][j] = c[i-1][j];
+            else c[i][j] = c[i][j-1];
+        }
+    }
+}
+
+void LCS(const Matrix& c){
+    // print all the path from bottom to up by dfs
+    if(c.size() == 0) return;
+    int m = c[0].size(), r = c.size(); // columns & rows
+    int i(r),j(m);
+    
+}
+
+void printMatrix(const Matrix& c){
+    for (int i=0; i<c.size(); i++) {
+        for (int j=0; j<c[i].size(); j++) {
+            cout << c[i][j] << ' ';
+        }
+        cout << endl;
+    }        
+}
+
+
+
+
+
+
+
+
+
+

ms2011/Algorithm/lcs.exe

@@ -0,0 +1,10 @@
+Exception: STATUS_ACCESS_VIOLATION at eip=004014AE
+eax=00000001 ebx=00000004 ecx=00900338 edx=00000000 esi=00000000 edi=00000000
+ebp=0022CCC8 esp=0022CCA0 program=G:\career\code\Algorithm\lcs.exe, pid 2456, thread main
+cs=001B ds=0023 es=0023 fs=003B gs=0000 ss=0023
+Stack trace:
+Frame     Function  Args
+0022CCC8  004014AE  (0022CD1C, 0022CD18, 0022CCEC, 0022CD17)
+0022CD38  004012C7  (00000001, 61243964, 009000F8, 61004BB9)
+0022CD68  61007038  (00000000, 0022CDA4, 61006980, 7FFD8000)
+End of stack trace

ms2011/Algorithm/lcs.exe.stackdump

@@ -0,0 +1,16 @@
+// 不用任何循环语句，库函数求一个输入字符串的长度
+
+#include <iostream>
+using namespace std;
+
+int lens(char *s){
+     if(*s == '\0') return 0;
+     else return 1+lens(++s);
+}
+
+int main(int argc,char *argv[])
+{
+    char * s = argv[1];
+    cout << lens(s) << endl;
+    return 0;
+}

ms2011/BS/aly1.cpp

@@ -0,0 +1,80 @@
+// dianp1.cpp
+// 大众点评第1题
+// 描述：
+//      Input:  一个n个数的数组,其中某个数字出现次数大于N/2;
+//      Outpu:  找出那个数
+//              
+
+#include <iostream>
+#include <vector>
+using namespace std;
+
+int findk(vector<int> a); // 计数法，连续相同则加，否则减
+int rfindk(vector<int> a,int s,int e); // 快排划分法,传值,以中间值划分的方式不适合
+int Partition(vector<int> &A, int s, int e );
+void QuickSort2( vector<int>& A, int p, int q );
+
+int main()
+{
+    int a[] = {1,2,3,2,2,2,5}; // test
+    vector<int> v(a,a+7);
+    cout << findk(v) << endl;
+    cout << rfindk(v,0,7) << endl;
+    return 0;
+}
+
+int findk(vector<int> a){
+    if (a.size() < 0) return -1;   //设置记录找不到的状态;
+    int i,vn(a[0]),cnt(1); 
+    for (i=1; i< a.size(); i++) {        
+        if(a[i] == a[i-1]) cnt ++;           
+        else cnt --;    
+        if(cnt == 0) vn = a[i];        
+    }
+    return vn;
+}
+
+int rfindk(vector<int> a,int s,int e){
+    int r = Partition(a,s,e);
+    if (e-s > a.size()/2) return a[(s+e)/2];
+    if (e-r > r-s) rfindk(a,r+1,e);
+    else rfindk(a,s,r-1);
+}
+
+int Partition(vector<int> &A, int s, int e )
+{// Asending 
+    int p_pos,tmp,i,j;//中间点位置    
+    p_pos = s; //假设中间点位置在第一位
+    //假设中间点值为第一位元素的值 注意中间点元素的值是确定的 假设第一位 但是中间点元素的正确位置需要我们找
+    int pivot = A[p_pos];
+    for(i=s+1; i<=e; i++){ //从第二位开始查找 发现i位元素小于中间点值
+        if(A[i] < pivot){
+          p_pos++; //中间点位置右移一位        
+          if(p_pos != i) swap<int>(A[p_pos],A[i]);//交换此刻中间点位置所在元素与第i位元素的值
+        }
+    }
+    //此时已经找到中间点元素所在正确位置 将中间点元素位置的值与假设第一位为中间点值交换 
+    swap<int>(A[s],A[p_pos]);       
+    return p_pos; //返回该中间点位置
+}
+
+void QuickSort2( vector<int>& A, int p, int q )
+{
+    if( p < q )
+    {
+        int r = Partition(A, p, q);
+        QuickSort2(A,p,r-1);
+        QuickSort2(A,r+1,q);
+    }
+}
+
+
+
+
+
+
+
+
+
+
+

ms2011/BS/aly1.exe

@@ -0,0 +1,16 @@
+Exception: STATUS_ACCESS_VIOLATION at eip=61111AC8
+eax=00A80478 ebx=FFFFFFF8 ecx=3FFF811D edx=00000000 esi=00AA0000 edi=00A9FFFC
+ebp=0022CBE8 esp=0022CBDC program=G:\career\code\BS\hulu.exe, pid 5740, thread main
+cs=001B ds=0023 es=0023 fs=003B gs=0000 ss=0023
+Stack trace:
+Frame     Function  Args
+0022CBE8  61111AC8  (00A80478, 00A8047C, FFFFFFF8, 611289CB)
+0022CC18  00401D26  (00A8047C, 00A80474, 00A80478, 00402369)
+0022CC48  004023A3  (00A8047C, 00A80474, 00A80478, 00402341)
+0022CC78  0040241D  (00A8047C, 00A80474, 00A80478, 00A80474)
+0022CC98  004025AA  (00A8047C, 00A80474, 00A80478, 00401B57)
+0022CCC8  00402218  (0022CCF0, 00A80478, 0022CCE8, 610C9BBA)
+0022CD08  00401281  (00000001, 61180E48, 00000000, 0022D000)
+0022CD38  0040132C  (00000001, 61243824, 00A800F8, 61004BB9)
+0022CD68  61007038  (00000000, 0022CDA4, 61006980, 7FFDF000)
+End of stack trace

ms2011/BS/dianp1.cpp

@@ -0,0 +1,43 @@
+// hulu1 笔试题- 50个数每次删除奇数/偶数最后剩下的数
+// by wrchow at 20110926
+
+#include <iostream>
+#include <vector>
+using namespace std;
+
+int oddeven50(int f){
+// 矩阵思想
+    if(f == 1) cout << "Delete odd.." << endl;
+    else cout << "Delete even.." << endl;
+    vector<int> v;
+    v.clear();
+    int i=0;
+    for (i=1; i<=50; i++) {
+        v.push_back(i);
+    }
+    while (v.size() != 1) { // 数组不能循环删除
+        vector<int> t;
+        for (i=f; i<v.size(); i+=2){   
+            t.push_back(v[i]);
+            cout << v[i] << ' ';
+        }  cout << endl;
+        v.clear();
+        v=t;
+    }   
+    return v[0];
+}
+
+int * postOrder() {
+    
+}
+
+int main()
+{
+    // ---------------------
+    cout << "odd & even of 1..50 is :"  // 32 & 1
+         << "odd = " <<  oddeven50(1)   
+         << ", even = " << oddeven50(0) << endl;  //后边的先调用  
+    // --------------------- 
+     
+    return 0;
+}