常见面试的查找和排序算法

短命女 2022-03-11 06:30 172阅读 0赞

# 一、排序总结 #

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（1）	快排
    	private void quicksort(int[] array, int begin, int end) {
    		// TODO Auto-generated method stub
    		if(begin<end)
    		{
    			int key=array[begin]; //设置第一个元素为基准元素
    			int i=begin;
    			int j=end;
    			
    			while(i<j)
    			{
    			  while(i<j&&array[j]>key)
    			  {
    				  j--;
    			  }
    			  if(i<j)
    			  {
    				  array[i]=array[j];
    				  i++;
    			  }
    			  while(i<j&&array[i]<key)
    			  {
    				  i++;
    			  }
    			  if(i<j)
    			  {
    				  array[j]=array[i];
    				  j--;
    			  }
    			  array[i]=key;  //放置基准元素
    			  
    			  
    			  quicksort(array,begin,i-1);
    			  
    			  quicksort(array,i+1,end);
    			  
    			}
    			
    			
    		}
    		
    	}

（2）	堆排序
    public static void heapSort(int[] arr) {
    		if (arr == null || arr.length < 2) {
    			return;
    		}
    		for (int i = 0; i < arr.length; i++) {
    			heapInsert(arr, i);
    		}
    		int size = arr.length;
    		swap(arr, 0, --size);
    		while (size > 0) {
    			heapify(arr, 0, size);
    			swap(arr, 0, --size);
    		}
    	}
    //建堆
    	public static void heapInsert(int[] arr, int index) {
    		while (arr[index] > arr[(index - 1) / 2]) {
    			swap(arr, index, (index - 1) / 2);
    			index = (index - 1) / 2;
    		}
    	}
    //堆调整
    	public static void heapify(int[] arr, int index, int size) {
    		int left = index * 2 + 1;
    		while (left < size) {
    			int largest = left + 1 < size && arr[left + 1] > arr[left] ? left + 1 : left;
    			largest = arr[largest] > arr[index] ? largest : index;
    			if (largest == index) {
    				break;
    			}
    			swap(arr, largest, index);
    			index = largest;
    			left = index * 2 + 1;
    		}
    	}
    //交换数据
    	public static void swap(int[] arr, int i, int j) {
    		int tmp = arr[i];
    		arr[i] = arr[j];
    		arr[j] = tmp;
    	}

（3）	归并排序
    public static void mergeSort(int[] arr) {
    		if (arr == null || arr.length < 2) {
    			return;
    		}
    		mergeSort(arr, 0, arr.length - 1);
    	}
    //归并具体操作
    	public static void mergeSort(int[] arr, int l, int r) {
    		if (l == r) {
    			return;
    		}
    		int mid = l + ((r - l) >> 1);
    		mergeSort(arr, l, mid);
    		mergeSort(arr, mid + 1, r);
    		merge(arr, l, mid, r);
    	}
    //合并过程
    	public static void merge(int[] arr, int l, int m, int r) {
    		int[] help = new int[r - l + 1];
    		int i = 0;
    		int p1 = l;
    		int p2 = m + 1;
    		while (p1 <= m && p2 <= r) {
    			help[i++] = arr[p1] < arr[p2] ? arr[p1++] : arr[p2++];
    		}
    		while (p1 <= m) {
    			help[i++] = arr[p1++];
    		}
    		while (p2 <= r) {
    			help[i++] = arr[p2++];
    		}
    		for (i = 0; i < help.length; i++) {
    			arr[l + i] = help[i];
    		}
    	}

# 二、查找总结 #

**顺序查找**:不需要数列本身有序，查找性能太差，平均查找长度为(1+n)/2,时间复杂度为O(n).  
**二分查找**: 要求数列有序  
**行列递增的矩阵查找**：按照杨氏矩阵特殊的存储方式  
**分块查找**：顺序查找和二分查找的改进  
其他查找算法:\*\*深度优先查找、广度优先查找、二叉树查找(先、中、后、层序遍历)\*\*等。

(1)	二分查找
    /**
    	 * 递归实现的二分查找
    	 * @param key
    	 */
    	public int SearchRecursion(int key)
    	{
    		if(array!=null)
    		{
    			return searchRecursion(key,0,array.length-1);
    		}
    		
    		return -1;
    	}
    	
    	private int searchRecursion(int key, int start, int end) 
    	{
    		// TODO Auto-generated method stub
    		if(start>end)
    		{
    			return -1;
    		}
    	    int mid=start+(end-start)/2;
    	    if(array[mid]==key)
    	    {
    	    	return mid;
    	    }
    	    else if(array[mid]<key)
    	    {
    	    	return searchRecursion( key, mid+1, end);
    	    }else
    	    {
    	    	return searchRecursion( key, start, mid-1);
    	    }
     
    	}
    	
    	
    
    
    
    
    
    	 /**
    	  * 二分查找的非递归实现
    	  * @param key
    	  */
    	public int search(int key)
    	{
    		if(array==null)
    		{
    			return -1;
    		}
    		int start=0;
    		int end=array.length-1;
    		while(start<=end)
    		{
    			int mid =start+(end-start)/2;
    			if(array[mid]==key)
    			{
    				return mid;
    			}
    			else if(array[mid]<key)
    			{
    				start=mid+1;
    			}else
    			{
    				end=mid-1;
    			}
    			
    		}
    		
    		return -1;
    	}

# 三、字符串匹配KMP #

经典KMP 算法
    
    //方法:获得模式串在母串的位置
       public static int KMP(String s,String m){
            if(s==null||m==null||m.length()<1||s.length()<m.length()){
                return -1;
            }
            char[]ss=s.toCharArray();  //主串字符串转数组
            char[]ms=m.toCharArray();  //子串字符串转数组
            int si=0;
            int mi=0;
            int[]next=getNextArray(ms);
            while(si<ss.length&&mi<ms.length){
                if(ss[si]==ms[mi]){
                    si++;
                    mi++;
                }else if(next[mi]==-1){
                    si++;
                }else{
                    mi=next[mi];
                }
            }
            return mi==ms.length?si-mi:-1;
    
        }
        //获得模式串的next数组
        public static int[]getNextArray(char[]ms){
           if(ms.length==1){
              return new int[]{-1};
           }
           int[]next=new int[ms.length];
           next[0]=-1;
           next[1]=0;
           int pos=2;
           int cn=0;
           while(pos<ms.length){
                if(ms[pos-1]==ms[cn]){
                    next[pos++]=++cn;
                }else if(cn>0){
                     cn=next[cn];
                }else{
                    next[pos++]=0;
                }
           }
           return next;
        }

# 四、二叉树的遍历 #

public static class Node {
            public int value;
            public Node left;
            public Node right;
    
            public Node(int data) {
                this.value = data;
            }
        }
       //先序遍历(递归)
        public static void preOrderRecur(Node head) {
            if (head == null) {
                return;
            }
            System.out.print(head.value + " ");
            preOrderRecur(head.left);
            preOrderRecur(head.right);
        }
       //中序遍历(递归)
        public static void inOrderRecur(Node head) {
            if (head == null) {
                return;
            }
            inOrderRecur(head.left);
            System.out.print(head.value + " ");
            inOrderRecur(head.right);
        }
       //后序遍历(递归)
        public static void posOrderRecur(Node head) {
            if (head == null) {
                return;
            }
            posOrderRecur(head.left);
            posOrderRecur(head.right);
            System.out.print(head.value + " ");
        }
    
        //先序遍历(非递归)
        public static void preOrderUnRecur(Node head) {
            System.out.print("pre-order: ");
            if (head != null) {
                Stack<Node> stack = new Stack<Node>();
                stack.add(head);
                while (!stack.isEmpty()) {
                    head = stack.pop();
                    System.out.print(head.value + " ");
                    if (head.right != null) {
                        stack.push(head.right);
                    }
                    if (head.left != null) {
                        stack.push(head.left);
                    }
                }
            }
            System.out.println();
        }
       //中序遍历(非递归)
        public static void inOrderUnRecur(Node head) {
            System.out.print("in-order: ");
            if (head != null) {
                Stack<Node> stack = new Stack<Node>();
                while (!stack.isEmpty() || head != null) {
                    if (head != null) {
                        stack.push(head);
                        head = head.left;
                    } else {
                        head = stack.pop();
                        System.out.print(head.value + " ");
                        head = head.right;
                    }
                }
            }
            System.out.println();
        }
            //后序遍历(非递归)
        public static void posOrderUnRecur2(Node head) {
            System.out.print("pos-order: ");
            if (head != null) {
                Stack<Node> s1 = new Stack<Node>();
                Stack<Node> s2 = new Stack<Node>();
                s1.push(head);
                while (!s1.isEmpty()) {
                    head = s1.pop();
                    s2.push(head);
                    if (head.left != null) {
                        s1.push(head.left);
                    }
                    if (head.right != null) {
                        s1.push(head.right);
                    }
                }
                while (!s2.isEmpty()) {
                    System.out.print(s2.pop().value + " ");
                }
            }
            System.out.println();
        }
       //后序遍历(非递归)
        public static void posOrderUnRecur2(Node h) {
            System.out.print("pos-order: ");
            if (h != null) {
                Stack<Node> stack = new Stack<Node>();
                stack.push(h);
                Node c = null;
                while (!stack.isEmpty()) {
                    c = stack.peek();
                    if (c.left != null && h != c.left && h != c.right) {
                        stack.push(c.left);
                    } else if (c.right != null && h != c.right) {
                        stack.push(c.right);
                    } else {
                        System.out.print(stack.pop().value + " ");
                        h = c;
                    }
                }
            }
            System.out.println();
    }
    
    //层序遍历
    public  static  void  levelTraverse(Node<T> root){
          if(root == null)
              return; 
    Queue<Node<T>> queue = new LinkedList<>();//层序遍历时保存结点的队列
            queue.offer(root);//初始化
            while(!queue.isEmpty()){
                BinaryNode<T> node = queue.poll();
                System.out.print(node.element + " ");//访问节点
                if(node.left != null)
                    queue.offer(node.left);
                if(node.right != null)
                    queue.offer(node.right);
    }
     
     }

# 五、图的遍历 #

(1)BFS
    public static void bfs(Node node) {
    		if (node == null) {
    			return;
    		}
    		Queue<Node> queue = new LinkedList<>();
    		HashSet<Node> map = new HashSet<>();
    		queue.add(node);
    		map.add(node);
    		while (!queue.isEmpty()) {
    			Node cur = queue.poll();
    			System.out.println(cur.value);
    			for (Node next : cur.nexts) {
    				if (!map.contains(next)) {
    					map.add(next);
    					queue.add(next);
    				}
    			}
    		}
    	}

(2)	DFS
    public static void dfs(Node node) {
    		if (node == null) {
    			return;
    		}
    		Stack<Node> stack = new Stack<>();
    		HashSet<Node> set = new HashSet<>();
    		stack.add(node);
    		set.add(node);
    		System.out.println(node.value);
    		while (!stack.isEmpty()) {
    			Node cur = stack.pop();
    			for (Node next : cur.nexts) {
    				if (!set.contains(next)) {
    					stack.push(cur);
    					stack.push(next);
    					set.add(next);
    					System.out.println(next.value);
    					break;
    				}
    			}
    		}
    	}

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