面试题目——《CC150》中等难题

面试题17.1：编写一个函数，不用临时变量，直接交换两个数。

　　思路：使用差值或者异或

package cc150.middle;

public class Exchange {

	public static void main(String[] args) {
		// TODO 自动生成的方法存根
		int[] a= {1,2};
		Exchange ec = new Exchange();
		int[] b= ec.exchangeAB(a);
		System.out.println(b[0]);
		System.out.println(b[1]);
	}

	 public int[] exchangeAB(int[] AB) {
	        // write code here
//		 	AB[0] = AB[0]-AB[1];
//		 	AB[1] = AB[0]+AB[1];
//		 	AB[0] = AB[1]-AB[0]; 

		 	AB[0] = AB[0]^AB[1];
		 	AB[1] = AB[0]^AB[1];
		 	AB[0] = AB[1]^AB[0];
		 	return AB;
	    }

}

面试题17.2：设计一个算法，判断玩家是否赢了井字游戏。

package cc150.middle;

public class Board {	//井字棋

	public static void main(String[] args) {
		// TODO 自动生成的方法存根
		int[][] a = {{1,0,1},{1,-1,-1},{1,-1,0}};
		Board b = new Board();
		System.out.println(b.checkWon(a));
	}

	public boolean checkWon(int[][] board) {
        // write code here
		int N = board.length;
		int row = 0;
		int col = 0;

		//检查行
		for(row = 0;row<N;row++){
			for(col = 1;col<N;col++){
				if(board[row][col] != board[row][col-1])
					break;
			}
			if(col == N && board[row][col-1] == 1)
				return true;
		}

		//检查列
		for(col = 0;col<N;col++){
			for(row = 1;row<N;row++){
				if(board[row][col] != board[row-1][col])
					break;
			}
			if(row == N && board[row-1][col] == 1)
				return true;
		}

		//检查对角线，从左上角开始
		if(board[0][0] != 0){
			for(row = 1;row<N;row++){
				if(board[row][row] != board[row-1][row-1])
					break;
			}
			if(row == N && board[row-1][row-1] == 1)
				return true;
		}

		//检查对角线，左下角开始
		if(board[N-1][0] != 0){
			for(row = 1;row<N;row++){
				if(board[N-row-1][row] != board[N-row][row-1])
					break;
			}
			if(row == N && board[N-row][row-1] == 1)
				return true;
		}
		return false;
    }

}

面试题17.3：设计一个算法，算出n阶乘有多少个尾随零。

package cc150.middle;

public class Factor {

	public static void main(String[] args) {		//n阶乘有多少个尾随零
		// TODO 自动生成的方法存根
		Factor f = new Factor();
		System.out.println(f.getFactorSuffixZero(10));
	}

	public int getFactorSuffixZero(int n) {//统计所有相乘的数中分解出5的个数，2的个数必定大于5，只要统计5的个数即可
        // write code here
//		int count = 0;
//		for(int i=2;i<=n;i++)
//			count += factorOf5(i);
//		return count;

		int count = 0;
		for(int i=5;n/i>0;i*=5)		//如果n=10的话，有2个数在5和25之间
			count += n/i;
		return count;
    }

	public int factorOf5(int n) {
        // write code here
		int count = 0;
		while(n%5 == 0){	//求余数
			count ++;
			n/=5;
		}
		return count;
    }

}

面试题17.4：编写一个方法，找出两个数字中最大的那一个。不得使用if-else或其他比较运算符。

package cc150.middle;

public class Max {

	public static void main(String[] args) {
		// TODO 自动生成的方法存根

	}

	public int getMax(int a, int b) {
        // write code here
		b = a-b;					//此时b>>31为1则b小于0即a<b,若b>>31为0 则a>b
	    a -= b&(b>>31); 	//若a<b a=a-(a-b),若a>b a=a-0
	    return a;
    }

}

面试题17.6：给定一个整数数组，编写一个函数，找出索引m和n，只要将m和n之间的元素排好序，整个数组就是有序的。注意：n-m越小越好，也就是说，找出符合条件的最短序列。

面试题17.7：给定一个整数，打印该整数的英文描述（例如“One Thousand,Two Hundred Thirty Four”）。

package cc150.middle;

public class ToString {

	public static void main(String[] args) {
		// TODO 自动生成的方法存根
		ToString ts = new ToString();
		System.out.println(ts.toString(1000));		//输入1234，输出"One Thousand,Two Hundred Thirty Four"
	}

	public static String[] digits = {"One","Two","Three","Four","Five","Six","Seven","Eight","Nine"};
	public static String[] teens = {"Eleven","Twelve","Thirteen","Fourteen","Fifteen","Sixteen","Seventeen","Eighteen","Nineteen"};
	public static String[] tens = {"Ten","Twenty","Thirty","Forty","Fifty","Sixty","Seventy","Eighty","Ninety"};
	public static String[] bigs = {"","Thousand","Million"};

	public String toString(int x) {
        // write code here
		if(x == 0)
			return "Zero";
		else if(x < 0)
			return "Negative " + toString(-1*x);
		int count = 0;
		String str = "";
		while(x > 0){
			if(x % 1000 != 0){
				if(count == 0 || str == "")  //One Million后面不能有逗号
					str = toString100(x % 1000)+bigs[count]+str; //把count大的million放在前面，原来的str放在后面
				else
					str = toString100(x % 1000)+bigs[count]+","+str; //把count大的million放在前面，原来的str放在后面
			}
			x /= 1000;
			count ++;
		}
		return str.trim();
    }

	public String toString100(int x){
		String str = "";
		//转换百位数的地方
		if(x >= 100){
			str+=digits[x/100-1]+" Hundred ";
			x%=100;
		}
		//转换十位数的地方
		if(x >= 11 && x <= 19){
			return str+teens[x-11]+" ";
		}else if(x == 10 || x >= 20){
			str+=tens[x/10-1]+" ";
			x%=10;
		}
		//转换个位数的地方
		if(x >= 1 && x<=9){
			str+=digits[x-1]+" ";
		}
		return str;
	}

}

面试题17.8：给定一个整数数组（有正数有负数），找出总和最大的连续数列，并返回总和。（剑指Offer原题）

package jianzhiOffer;

public class FindGreatestSumOfSubArray {		//连续子数组的最大和

	public static void main(String[] args) {
		// TODO 自动生成的方法存根
//		int[] Arr = {-1,-2,-3,-10,-4,-7,-2,-5};
		int[] Arr = {6,-3,-2,7,-15,1,2,2};
		System.out.println(findGreatestSumOfSubArray(Arr));
	}

	public static boolean InvalidInput = false;

	public static int findGreatestSumOfSubArray(int[] nums){
		if(nums == null || nums.length<=0){
			InvalidInput = true;
			return 0;
		}
		InvalidInput = true;
		int curSum = 0;
		int curGreatestSum = Integer.MIN_VALUE;//数组内可能全部都是负数
		for(int i=0;i<nums.length;i++){
			if(curSum < 0)//如果前n个相加小于0，那么最大的和只能从n+1个开始算
				curSum = nums[i];
			else
				curSum += nums[i];
			if(curSum > curGreatestSum)//数组内可能全部都是负数
				curGreatestSum = curSum;
		}
		return curGreatestSum;
	}

}

面试题17.9：设计一个方法，找出任意指定单词在一本书中的出现频率。

package cc150.middle;

import java.util.Hashtable;

public class Frequency {

	public static void main(String[] args) {
		// TODO 自动生成的方法存根

	}

	public int getFrequency(String[] article, int n, String word) {
        // write code here
		return wordFrequency(setupDictionary(article),word);
    }

	public Hashtable<String,Integer> setupDictionary(String[] book){
		Hashtable<String,Integer> table = new Hashtable<String,Integer>();
		for(String word:book){
			word = word.toLowerCase();
			if(word.trim() != ""){		//去掉空格之后，不等于空
				if(!table.containsKey(word))
					table.put(word, 0);
				table.put(word, table.get(word)+1);
			}
		}
		return table;
	}

	public int wordFrequency(Hashtable<String,Integer> table, String word){
		if(table == null || word == null)
			return -1;
		word = word.toLowerCase();
		if(table.containsKey(word))
			return table.get(word);
		return 0;
	}

}

面试题17.12：设计一个算法，找出数组中两数之和为指定值的所有整数对。

package cc150.middle;

import java.util.Arrays;

public class FindPair {

	public static void main(String[] args) {		//有重复的情况
		// TODO 自动生成的方法存根
		int[] a = {11,7,7,6,14,2,14,15,2,1,2,12,13,9,8,15,13,8,10,11,14,10,2,9,4,9,3,7,6,10,15,4,7,6,15,3,9,13,5,2,6,10,10,1,12,4,3,3,8,8,1,4,7,11,13,5,13,15,4,3,1,11,6,11,9,9,11,15,12,10,13,3,11,4,8,9,7,3,13,9,11,3,2,11,10,1,4,2,3,3,14,11,5,10,1,14,8,1,11,3,1,9,14,6,1,7,15,10,14,6,4,12,11};
		FindPair fp = new FindPair();
		System.out.println(fp.countPairs(a,113,16));
	}

	public int countPairs(int[] A, int n, int sum) {
        // write code here
		Arrays.sort(A);
		for(int i=0;i<A.length;i++)
			System.out.println(A[i]);
		int first = 0;
		int end = n-1;
		int count=0;
		int countLeft = 1;
		int countRight = 1;
		while(first<end){
			int s = A[first]+A[end];
			if(s == sum){							//如果相等
				if(A[first] == A[end]){		//情况1，一串连续的，且这些相加等于sum
					count += ((end-first)*(end-first+1)/2);
					return count;		//注意要在这里结束，否则会继续计数
				}else{									//情况2，两串连续的，两个串中各取一个相加等于sum
					while(A[first] == A[first+1]){
						countLeft++;
						first++;
					}
					while(A[end] == A[end-1]){
						countRight++;
						end--;
					}
					count += (countLeft*countRight);
				}
				first++;
				end--;
				countLeft=1;		//注意开始计数是1
				countRight=1;
			}else{
				if(s>sum)
					end--;
				else
					first++;
			}
		}
		return count;
    }

}

面试题17.13：有个简单的类似结点的数据结构BiNode，包含两个指向其他结点的指针。数据结构BiNode可用来表示二叉树（其中node1为左子节点，node2为右子节点）或双向链表（其中node1为前趋结点，node2为后继结点）。编写一个方法，将二叉查找树（用BiNode实现）转换为双向链表。要求所有数值的排序不变，转换操作不得引入其他数据结构（即直接操作原先的数据结构）。

package cc150.middle;

public class Converter {

	public static void main(String[] args) {		//二叉查找树转换成双向链表
		// TODO 自动生成的方法存根
		Converter cv = new Converter();
		TreeNode t4 = cv.new TreeNode(4);
		TreeNode t2 = cv.new TreeNode(2);
		TreeNode t5 = cv.new TreeNode(5);
		TreeNode t1 = cv.new TreeNode(1);
		TreeNode t3 = cv.new TreeNode(3);
		TreeNode t6 = cv.new TreeNode(6);
		TreeNode t0 = cv.new TreeNode(0);
		t4.left = t2;
		t4.right = t5;
		t2.left = t1;
		t2.right = t3;
		t5.right = t6;
		t1.left = t0;
		cv.treeToList(t4);
		while(cv.head.next != null){
			System.out.println(cv.head.val);
			cv.head = cv.head.next;
		}

	}

	private ListNode head  = new ListNode(-1);
	private ListNode q  = head;

	public ListNode treeToList(TreeNode root) {		//使用中序遍历
        // write code here
		if(root != null){
			treeToList(root.left);
			q.next = new ListNode(root.val);
			q = q.next;
			treeToList(root.right);
		}
		return head.next;

    }

	public class ListNode {
	    int val;
	    ListNode next = null;

	    ListNode(int val) {
	        this.val = val;
	    }
	}

	public class TreeNode {
	    int val = 0;
	    TreeNode left = null;
	    TreeNode right = null;
	    public TreeNode(int val) {
	        this.val = val;
	    }
	}

}