先浅尝一个汉诺塔(已刷2)

public class Solution{public static void main(String args[]){hanoi(2, 'A', 'B', 'C');}public static void hanoi(int n, char a, char b, char c) {if (n == 1) {System.out.println("第1块从"+ a + "到"+ c);}else {hanoi(n-1, a, c, b);System.out.println("第"+n+"块从"+ a + "到"+ c);hanoi(n-1, b, a, c);}}
}

合并两个有序链表（刷2）

题目

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode mergeTwoLists(ListNode list1, ListNode list2) {if (list1 == null) {return list2;}else if(list2 == null) {return list1;}else if(list1.val <= list2.val){// 这里其实就是一个概括性的步骤，我们并不去仔细想调用自身又干了什么// 概括性的步骤意思就是： 如果我当前l1的值小于l2的值，那说明当前//l1一定是在l1剩下的和整个l2完成合并之前list1.next = mergeTwoLists(list1.next, list2);return list1;}else{list2.next = mergeTwoLists(list1, list2.next);return list2;}}
}

移除链表元素(刷2)

题目

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode removeElements(ListNode head, int val) {if (head == null) {return null;}else if (head.val == val && head.next == null) {return null;}else if(head.val == val && head.next != null) {return removeElements(head.next, val);}else{head.next = removeElements(head.next, val);return head;}}
}

反转链表(刷2)

题目

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode reverseList(ListNode head) {if(head == null || head.next == null) {return head;}else {ListNode cur = reverseList(head.next);head.next.next = head;head.next = null;return cur;}}
}

2 的幂

题目

class Solution {public boolean isPowerOfTwo(int n) {if(n == 0){return false;}else if (n == 1){return true;}else if (n % 2 == 0 && isPowerOfTwo(n/2)) {   return true;}else {return false;}}
}

3 的幂

题目

class Solution {public boolean isPowerOfThree(int n) {if(n == 0 || n == 2) {return false;}else if (n == 1) {return true;} else if (n % 3 == 0 && isPowerOfThree(n/3)) {   return true;}else {return false;}}
}

4 的幂

题目

class Solution {public boolean isPowerOfFour(int n) {// 方法一，除到底//     while(n != 0 && n % 4 == 0) {//         n = n / 4;//     }//     return n == 1;// 方法二， 递归if (n < 4 && n != 1) {// 0 1 2 3return false;}else if (n == 1) {return true;}else if(n % 4 == 0 && isPowerOfFour(n/4)){return true;}else{return false;}}}

反转字符串

题目

class Solution {char t = 0;public void reverseString(char[] s) {reverse(s, 0, s.length - 1);}public void reverse(char[] s,int begin,int end) {if (end - begin >= 1) {// 交换首位t = s[end];s[end] = s[begin];s[begin] = t;reverse(s, begin+1, end - 1);}else {return;}}}

进阶版

1. 两数相加

题目

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode addTwoNumbers(ListNode l1, ListNode l2) {return r(l1,l2,0);}private ListNode r(ListNode l1,ListNode l2,int add){if(l1 == null && l2 == null && add == 0) {return null;}int val = (l1 == null? 0: l1.val) + (l2==null ? 0: l2.val) + add;add = val >= 10 ? 1 : 0;ListNode x = new ListNode(val % 10);x.next = r(l1==null?null:l1.next,l2==null?null:l2.next, add);return x;}}

2. 重排链表

题目

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public void reorderList(ListNode head) {if(head.next == null || head.next.next == null) {return;}else{ListNode end = getEnd(head);ListNode end2 = get2End(head);end.next = head.next;head.next = end;end2.next = null;reorderList(head.next.next);}}public ListNode getEnd(ListNode head) {// 给定一个链表得到他的尾指针while(head.next != null){head = head.next;}return head;}public ListNode get2End(ListNode head) {// 给定一个链表得到他的倒数第二指针while(head.next.next != null){head = head.next;}return head;}
}

一种非常好的思路

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public void reorderList(ListNode head) {if(head.next == null || head.next.next == null) {return;}else{List<ListNode> l = new ArrayList<ListNode>();ListNode cur = head;while(cur != null) {l.add(cur);cur = cur.next;}int i = 0;int j = l.size() - 1;while(i < j) {l.get(i).next = l.get(j);i++;if(i == j) {break;}else{l.get(j).next = l.get(i);j--;}}l.get(i).next = null;}}}