【剑指OFFER-二刷】

反转链表

输入一个链表，反转链表后，输出链表的所有元素。

不借助任何多余空间，代码如下：

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

ListNode(int val) {
this.val = val;
}
}*/
public class Solution {
public ListNode ReverseList(ListNode head) {
if(head == null)
return null;

ListNode left = head;
ListNode current = head;
ListNode right = head.next;

while(current.next != null) {
current.next = right.next;
right.next = left;
left = right;
right = current.next;
}
return left;
}
}

重建二叉树

输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

一定要注意确定前后界限，可以用个例子去试。

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
if(pre == null || in == null)
return null;

return reConstructBinaryTree(pre, in, 0, pre.length, 0, in.length);
}
public TreeNode reConstructBinaryTree(int[] pre, int[] in, int preS, int preE, int inS, int inE) {
if(preE <= preS || inE <= inS)
return null;

int i = inS;
int count = 0;
for(; i < inE && in[i] != pre[preS]; i++) {count++;}

TreeNode root = new TreeNode(pre[preS]);
root.left = reConstructBinaryTree(pre, in, preS+1, preS+1+count, inS, inS+count);
root.right = reConstructBinaryTree(pre, in, preS+1+count, preE, inS+count+1, inE);

return root;
}
}

字符串的排列

回溯法实现全排列。这里利用set实现去重的目的。

import java.util.ArrayList;
import java.util.*;

public class Solution {
public ArrayList<String> Permutation(String str) {
ArrayList<String> result = new ArrayList<>();
if(str == null || str.length() == 0)
return result;
HashSet<String> set = new HashSet<>();
Permutation(set, str.toCharArray(), 0);
result.addAll(set);
Collections.sort(result);
return result;
}

public void Permutation(HashSet<String> set, char[] array, int k) {
if(k > array.length - 1) {
set.add(new String(array));
return;
}
for(int i = k; i < array.length; i++) {
swap(array, k, i);
Permutation(set, array, k+1);
swap(array, k, i);
}
}

public void swap(char[] array, int a, int b) {
char tmp = array[a];
array[a] = array[b];
array[b] = tmp;
}
}

数组中只出现一次的数字

//num1,num2分别为长度为1的数组。传出参数
//将num1[0],num2[0]设置为返回结果
public class Solution {
public void FindNumsAppearOnce(int [] array,int num1[] , int num2[]) {
if(array == null || array.length <= 1) {
num1[0] = 0;
num2[0] = 0;
return;
}
int diff = array[0];
for(int i = 1; i < array.length; i++)
diff ^= array[i];

String str = Integer.toBinaryString(diff);
int index = str.length() - 1;
for(; index >= 0; index--) {
if(str.charAt(index) != '0')
break;
}
index = str.length() - 1 - index;

for(int i = 0; i < array.length; i++) {
if(isBitZero(array[i], index))
num1[0] ^= array[i];
else
num2[0] ^= array[i];
}
}

public boolean isBitZero(int n, int index) {
int assi = 1;
for(int i = 0; i < index; i++)
assi <<= 1;
return ((assi & n) == 0) ? true : false;
}
}

数组中重复的数字

这个解法时间复杂度O(N)，不新建空间。很巧妙。

public class Solution {
// Parameters:
//    numbers:     an array of integers
//    length:      the length of array numbers
//    duplication: (Output) the duplicated number in the array number,length of duplication array is 1,so using duplication[0] = ? in implementation;
//                  Here duplication like pointor in C/C++, duplication[0] equal *duplication in C/C++
//    这里要特别注意~返回任意重复的一个，赋值duplication[0]
// Return value:       true if the input is valid, and there are some duplications in the array number
//                     otherwise false
public boolean duplicate(int numbers[],int length,int [] duplication) {
if(numbers == null || numbers.length == 0 || length == 0)
return false;
for(int i = 0; i < length; i++) {
int index = numbers[i];
if(index < 0)
index += length;
if(numbers[index] < 0) {
duplication[0] = index;
return true;
}
else
numbers[index] = numbers[index] - length;

}
return false;
}
}

链表中环的入口结点

下面的解法很巧妙，遍历环的时候把上面的结点与这个结点解锁，这样最后一个没有next的就是环的入口节点。

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

ListNode(int val) {
this.val = val;
}
}
*/
public class Solution {

public ListNode EntryNodeOfLoop(ListNode pHead)
{
if(pHead == null || pHead.next == null)
return null;
while(pHead.next != null) {
ListNode p = pHead.next;
pHead.next = null;
pHead = p;
}
return pHead;
}
}