算法笔试题总结

一、递归和分制策略汉诺塔递归实现：Void hannoi(int n,char x,char y,char z){if (n ==1)move(x,1,z);//printf(“%i Move disk %i from %c to %c \n”,++c,n,x,z)//将编号1从x移到zelse{hanoi(n-1, x,z,y);//将x上编号1至n-1的移到y，z做辅助move(x,n,z);// 将编号n从x移到zhanoi(n-1, y,x,z);// 将y上编号1至n-1的移到z，x做辅助}}递归小结优点：结构清晰，可读性强，而且容易用数学归纳法来证明算法的正确性，因此它为设计算法、调试程序带来很大方便。

缺点：递归算法的运行效率较低，无论是耗费的计算时间还是占用的存储空间都比非递归算法要多。

解决方法：在递归算法中消除递归调用，使其转化为非递归算法。

1.采用一个用户定义的栈来模拟系统的递归调用工作栈。

该方法通用性强，但本质上还是递归，只不过人工做了本来由编译器做的事情，优化效果不明显。

2.用递推来实现递归函数。

3.通过Cooper变换、反演变换能将一些递归转化为尾递归，从而迭代求出结果。

后两种方法在时空复杂度上均有较大改善，但其适用范围有限。

排序总结直接插入排序：void insertSort(listtp & r){ for(i=2; i<=n; i++){ r[0]=r[i]; /* 设置监视哨*/j=i-1; k=r[i].key;while(j>0&&r[j].key>k)/* 大值的记录后移*/ { r[j+1]=r[j]; j--; }r[j+1]=r[0]; /*插入位置*/}} /* straisort */希尔排序：将整个序列划分成若干子序列，分别进行直接插入排序。

void ShellSort(SeqList R,int dlta[],int t){For(k=0;k<t;k++)ShellInsert(R，dlta[k])； //一趟增量为dlta[k]的Shell插入排序} //ShellSort{ for(i=dk+1; i<=n; i++){ r[0]=r[i]; /* 设置监视哨*/j=i-dk; k=r[i].key;while( r[j].key>k)/* 大值的记录后移*/{ r[j+dk]=r[j]; j-=dk; }r[j+dk]=r[0]; /*插入位置*/}} /* straisort */冒泡排序：void buble_sort(listtp r){ for (i=1; i<=n-1; i++) /* 共计n-1趟 */for (j=1; j<=n-i ; j++)if (r[j].key>r[j+1].key){ x=r[j]; r[j]=r[j+1] r[j+1]=x;} } /* bubble_sort */快速排序：在快速排序中，记录的比较和交换是从两端向中间进行的，关键字较大的记录一次就能交换到后面单元，关键字较小的记录一次就能交换到前面单元，记录每次移动的距离较大，因而总的比较和移动次数较少。

private static void qSort(sqlist & L,int low, int high){if (low<high) { //int pivotloc=partition(L,low,high); //以a[pivotloc]为基准元素将a[low:high]划分成3段a[low: pivotloc -1],a[pivotloc]和a[pivotloc +1:high]，使得a[low: pivotloc-1]中任何元素小于等于a[pivotloc]，a[pivotloc +1:high]中任何元素大于等于a[pivotloc]。

下标pivotloc在划分过程中确定。

qSort (L,low, pivotloc-1); //对左半段排序qSort(L, pivotloc+1,high); //对右半段排序 }}Int Partition(sqlist &L,int low,int high){L.r[0]=L.r[low];//用子表第一个元素作为轴Pivotkey = L.r[low].key;While(low<high){While(low<high&&L.r[high].key>=privotkey )High--;If(L.r[high].key<privotkey)R[low]=r[high];While(low<high&&L.r[low].key<=privotkey)Low++;If(L.r[low].key>privotkey)R[high]=r[low];}L.r[low]=L.r[0];Return low;}归并排序：public static void mergeSort(Comparable a[], int left, int right){int m = (left+right)/2; //取中点if(left>=right)return;if(left+1==right){if(a[left]>a[right])std::swap(a[left], a[right]);return;}mergeSort(a, left, m);mergeSort(a, m+1, right);merge(a, left,m, right); //合并到数组a}}void Merge(A[], int l, int m, int h){int i = l;int j = m+1;int k = 0; while(i<=m&&j<=h){if(A[i]<A[j]){B[k++] = A[i];i++;}else{B[k++] = A[j];j++;}}while(i<=m)B[k++] = A[i++];while(j<=h)B[k++] = A[j++];for(i=l; i<=h; i++)A[i] = B[i-l];}二分法：#include<iostream>#define N 10using namespace std;int main(){int a[N],front,end,mid,x,i;cout<<"请输入已排好序的a数组元素:"<<endl;for(i=0;i<N;i++)cin>>a[i];cout<<"请输入待查找的数x:"<<endl;cin>>x;front=0;end=N-1;mid=(front+end)/2;while(front<=end&&a[mid]!=x){if(a[mid]<x)front=mid+1;if(a[mid]>x)end=mid-1;mid=(front+end)/2;}if(a[mid]!=x)cout<<"没找到！"<<endl;elsecout<<"找到了！在第"<<mid+1<<"项里。

"<<endl;return 0;}二叉树递归、非递归遍历package edu.cumt.jnotnull;import java.util.Stack;public class BinaryTree {protected Node root;public BinaryTree(Node root) {this.root = root;}public Node getRoot() {return root;}/** 构造树 */p ublic static Node init() {Node a = new Node('A');Node b = new Node('B', null, a);Node c = new Node('C');Node d = new Node('D', b, c);Node e = new Node('E');Node f = new Node('F', e, null);Node g = new Node('G', null, f);Node h = new Node('H', d, g);return h;// root}/** 访问节点 */p ublic static void visit(Node p) {System.out.print(p.getKey() + " ");}/** 递归实现前序遍历 */p rotected static void preorder(Node p) {if (p != null) {visit(p);preorder(p.getLeft());preorder(p.getRight());}}/** 递归实现中序遍历 */p rotected static void inorder(Node p) {if (p != null) {inorder(p.getLeft());visit(p);inorder(p.getRight());}}/** 递归实现后序遍历 */p rotected static void postorder(Node p) {if (p != null) {postorder(p.getLeft());postorder(p.getRight());visit(p);}}/** 非递归实现前序遍历 */protected static void iterativePreorder2(Node p) {Stack<Node> stack = new Stack<Node>();Node node = p;while (node != null || stack.size() > 0) {while (node != null) {//压入所有的左节点，压入前访问它visit(node);stack.push(node);node = node.getLeft();}if (stack.size() > 0) {//node = stack.pop();node = node.getRight();}}}/** 非递归实现后序遍历 */protected static void iterativePostorder(Node p) {Node q = p;Stack<Node> stack = new Stack<Node>();while (p != null) {// 左子树入栈for (; p.getLeft() != null; p = p.getLeft())stack.push(p);// 当前节点无右子或右子已经输出while (p != null && (p.getRight() == null || p.getRight() == q)) {visit(p);q = p;// 记录上一个已输出节点if (stack.empty())return;p = stack.pop();}// 处理右子stack.push(p);p = p.getRight();}}/** 非递归实现后序遍历双栈法 */p rotected static void iterativePostorder2(Node p) {Stack<Node> lstack = new Stack<Node>();Stack<Node> rstack = new Stack<Node>();Node node = p, right;do {while (node != null) {right = node.getRight();lstack.push(node);rstack.push(right);node = node.getLeft();}node = lstack.pop();right = rstack.pop();if (right == null) {visit(node);} else {lstack.push(node);rstack.push(null);}node = right;} while (lstack.size() > 0 ||rstack.size() > 0);}/** 非递归实现后序遍历单栈法*/p rotected static void iterativePostorder3(Node p) {Stack<Node> stack = new Stack<Node>();Node node = p, prev = p;while (node != null || stack.size() > 0) { while (node != null) {stack.push(node);node = node.getLeft();}if (stack.size() > 0) {Node temp = stack.peek().getRight();if (temp == null || temp == prev) {node = stack.pop();visit(node);prev = node;node = null;} else {node = temp;}}}}/** 非递归实现后序遍历4 双栈法*/p rotected static void iterativePostorder4(Node p) {Stack<Node> stack = new Stack<Node>();Stack<Node> temp = new Stack<Node>();Node node = p;while (node != null || stack.size() > 0) { while (node != null) {temp.push(node);stack.push(node);node = node.getRight();}if (stack.size() > 0) {node = stack.pop();node = node.getLeft();}}while (temp.size() > 0) {node = temp.pop();visit(node);}}/** 非递归实现中序遍历 */p rotected static void iterativeInorder2(Node p) {Stack<Node> stack = new Stack<Node>();Node node = p;while (node != null || stack.size() > 0) { while (node != null) {stack.push(node);node = node.getLeft();}if (stack.size() > 0) {node = stack.pop();visit(node);node = node.getRight();}}}/*** @param args*/p ublic static void main(String[] args) {BinaryTree tree = new BinaryTree(init());System.out.print(" Pre-Order:");preorder(tree.getRoot());System.out.println();System.out.print(" In-Order:");inorder(tree.getRoot());System.out.println();System.out.print("Post-Order:");postorder(tree.getRoot());System.out.println();System.out.print(" Pre-Order:");iterativePreorder(tree.getRoot());System.out.println();System.out.print("Pre-Order2:");iterativePreorder2(tree.getRoot());System.out.println();System.out.print(" In-Order:");iterativeInorder(tree.getRoot());System.out.println();System.out.print(" In-Order2:");iterativeInorder2(tree.getRoot());System.out.println();System.out.print(" Post-Order:");iterativePostorder(tree.getRoot());System.out.println();System.out.print("Post-Order2:");iterativePostorder2(tree.getRoot());System.out.println();System.out.print("Post-Order3:");iterativePostorder3(tree.getRoot());System.out.println();System.out.print("Post-Order4:");iterativePostorder4(tree.getRoot());System.out.println();}}字符串反转//字符串反转char *reverse(char *str){int len = strlen(str);char temp ;int i;char *str1;str1 = (char*)malloc(len);memset(str1,0,len);memcpy(str1,str,len);for( i=0; i <len/2; i++){temp = str1[i];str1[i] = str1[len-1-i];//*(str+len-1-i); //执行到该行出现系统错误str1[len-1-i]= temp;}return str1;}function reverse(arg) {if(arg.length == 0) {return arg;}else {return reverse(arg.substr(1,arg.length)) + arg.substr(0,1);}}alert(reverse("123456"));2009-01-23 00:26快速排序的原理倒是挺简单，选一个基准，将其余的元素与之对比分别放在左右两侧，左大右小，每次对比都是对其所有的元素，也就是说每一次都是n次对比。