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  • 剑指Offer学习笔记（2）

    一、代码的完整性

    1. 数值的整数次方
    给定一个double类型的浮点数base和int类型的整数exponent。求base的exponent次方。
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    publicclassSolution {
            publicdoublePower(doublebase, intexponent) {
            if(exponent==0){
                return1;
            }else  if(exponent>0){
                doubleresult=1;
                for(inti=1;i<=exponent;i++){
                    result*=base;
                }
                returnresult;
            }else{
                doubleresult=1.0; 
                exponent=-exponent;
                for(inti=1;i<=exponent;i++){
                    result*=(double)base;
                }
                 
                return((double)1.0)/((double)result);
            }
            //return exponent;
     
        }
    }

    2. 调整数组顺序使奇数位于偶数前面

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    publicclassSolution {
      publicvoidreOrderArray(int[] array) {
     
            intindexEven = -1;
     
            for(inti = 0; i < array.length; i++) {
                if(array[i] % 2== 1) {
                    if(indexEven < i && indexEven != -1) {
                        inttemp = array[i];
                        for(intj = i; j > indexEven; j--) {
                            array[j] = array[j-1];
                        }
                        array[indexEven] = temp;
                        indexEven += 1;
                    }
     
                } else{
                    if(indexEven == -1) {
                        indexEven = i;
                    }
                }
     
            }
     
        }
    }


     3. 反转链表

    输入一个链表，反转链表后，输出链表的所有元素。
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    /*
    public class ListNode {
        int val;
        ListNode next = null;
     
        ListNode(int val) {
            this.val = val;
        }
    }*/
    publicclassSolution {
        publicListNode ReverseList(ListNode head) {
            ListNode cur=head;
            ListNode next=null;
            ListNode pre=null;
            if(head==null||head.next==null){
                returnhead;
            }
            while(cur!=null){
                next=cur.next;
                cur.next=pre;
                 
                pre=cur;
                cur=next;
            }
            returnpre;
        }
    }
    4. 链表中倒数第k个结点
    代码思路如下：两个指针，先让第一个指针和第二个指针都指向头结点，然后再让第一个指正走(k-1)步，到达第k个节点。然后两个指针同时往后移动，当第一个结点到达末尾的时候，第二个结点所在位置就是倒数第k个节点了。
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    /*
    public class ListNode {
        int val;
        ListNode next = null;
     
        ListNode(int val) {
            this.val = val;
        }
    }*/
    publicclassSolution {
        publicListNode FindKthToTail(ListNode head, intk) {
     
            ListNode first = head;
            ListNode second = head;
     
            if(head == null||k<=0) {
                returnnull;
            }
            for(inti = 1; i < k; i++) {
                if(second.next != null) {
                    second = second.next;
                } else{
                    returnnull;
                }
            }
            while(second.next!=null){
                second=second.next;
                first=first.next;
            }
             
             
     
            returnfirst;
        }
     
    }
     
    5. 打印1到最大的n位数
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    publicclassSolution {
        /**
         * 递归调用的思路
         * @param n
         */
        publicstaticvoidPrint1ToMaxOfNDigits(intn) {
            if(n<=0){
                return;
            }
             
            char[] num = newchar[n];
            //num[n]='';
             
            for(inti=0;i<10;i++){
                num[0]=(char)(i+'0');//字符转化成数字
                Print1ToMaxOfNDigitsRecursive(num,n,0);
            }
             
     
        }
     
        privatestaticvoidPrint1ToMaxOfNDigitsRecursive(char[] num, intlength, intindex) {
            if(index==length-1){//如果组合完毕
    //          System.out.print(num);
                printNum(num);
                return;
            }
             
            for(inti=0;i<10;i++){
                num[index+1]=(char) ((char) i+'0');
                Print1ToMaxOfNDigitsRecursive(num, length, index+1);
            }
        }
     
        privatestaticvoidprintNum(char[] num) {
            booleanisBegin0 = true;
            intlen=num.length;
            for(inti=0;i<len;i++){
                if(isBegin0&&num[i]!='0'){
                    isBegin0=false;
                }
                 
                if(!isBegin0){
                    System.out.print(num[i]);
                }
                 
            }
            System.out.print('  ');
             
        }

    二、代码的完整性

    1. 链表中倒数第k个结点
    输入一个链表，输出该链表中倒数第k个结点。
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    /*
    public class ListNode {
        int val;
        ListNode next = null;
     
        ListNode(int val) {
            this.val = val;
        }
    }*/
    publicclassSolution {
        publicListNode FindKthToTail(ListNode head, intk) {
     
            ListNode first = head;
            ListNode second = head;
     
            if(head == null||k<=0) {
                returnnull;
            }
            for(inti = 1; i < k; i++) {
                if(second.next != null) {
                    second = second.next;
                } else{
                    returnnull;
                }
            }
            while(second.next!=null){
                second=second.next;
                first=first.next;
            }
             
             
     
            returnfirst;
        }
     
    }
    相关题目：找中间结点可以两个指针，一个一次走一个 一个一次走两个，快的首先到结尾。
    判断是否是环形链表，快的追上慢的。
    2. 反转链表
    输入一个链表，反转链表后，输出链表的所有元素。
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    /*
    public class ListNode {
        int val;
        ListNode next = null;
     
        ListNode(int val) {
            this.val = val;
        }
    }*/
    publicclassSolution {
            publicListNode ReverseList(ListNode head) {
            ListNode temp=null;
            ListNode orgin=null;
            if(head==null){
                returnhead;
            }
            temp=orgin=head;
            head=head.next;
            orgin.next=null;
            while(head!=null){ 
                temp=head;
                head=head.next;
                temp.next=orgin;
                orgin=temp;
            }
     
            returntemp;
        }
    }
    3. 反转链表
    输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。
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    /*
    public class ListNode {
        int val;
        ListNode next = null;
     
        ListNode(int val) {
            this.val = val;
        }
    }*/
    publicclassSolution {
        publicListNode Merge(ListNode list1,ListNode list2) {
            if(list1==null){
                returnlist2;
            }
            if(list2==null){
                returnlist1;
            }
            ListNode mergeList = null;
            if(list1.val<list2.val){
                mergeList=list1;
                mergeList.next=Merge(list1.next,list2);
            }else{
                mergeList=list2;
                mergeList.next=Merge(list1,list2.next);
            }
            returnmergeList;
        }
    }

    4. 树的子结构
    输入两颗二叉树A，B，判断B是不是A的子结构。
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    /**
    public class TreeNode {
        int val = 0;
        TreeNode left = null;
        TreeNode right = null;
     
        public TreeNode(int val) {
            this.val = val;
     
        }
     
    }
    */
    publicclassSolution {
        publicbooleanHasSubtree(TreeNode root1, TreeNode root2) {
            booleanflag=false;
            if(root1!=null&&root2!=null){
                if(root1.val==root2.val){
                    flag=doesTree1HaveTree2(root1,root2);
                }
                if(!flag){
                    flag=HasSubtree(root1.left,root2);
                }
                if(!flag){
                    flag=HasSubtree(root1.right,root2);
                }
            }
            returnflag;
        }
     
        privatebooleandoesTree1HaveTree2(TreeNode root1, TreeNode root2) {
            if(root2==null){
                returntrue;
            }
             
            if(root1==null){
                returnfalse;
            }
             
            if(root1.val!=root2.val){
                returnfalse;
            }
            if(root1.val==root2.val){
                returndoesTree1HaveTree2(root1.left, root2.left)&&doesTree1HaveTree2(root1.right, root2.right);
            }
            returnfalse;
        }
    }



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  • 原文地址：https://www.cnblogs.com/mrzhang123/p/5365800.html
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