【剑指offer】编程习题集附上答案

剑指offer~编程小哥令狐

一、数组类~

03、数组中重复的数字

class Solution
 {
    public void swap(int[] nums,int i,int j)
    {
        int temp=nums[i];
        nums[i]=nums[j];
        nums[j]=temp;
    }
    public int findRepeatNumber(int[] nums) 
    {
        int n=nums.length;
        for(int num:nums)
            if(num<0||num>=n)
                return -1;
        
        for(int i=0;i<n;i++)
        {
            while(nums[i]!=i&&nums[i]!=nums[nums[i]])
                swap(nums,i,nums[i]);
            if(nums[i]!=i&&nums[i]==nums[nums[i]])
                   return nums[i];
        }
         return -1;
    }
}

04、二维数组中的查找

• 暴力查找

class Solution {
    public boolean findNumberIn2DArray(int[][] matrix, int target) {
        int i,j;
        if((matrix==null||matrix.length==0)||(matrix.length==1&&matrix[0].length==0))
            return false;
        for (i=0;i<matrix.length;i++)
            for (j=0;j<matrix[0].length;j++)
                if(matrix[i][j]==target)
                    return true;
        
        return false;

    } 
}

• 线性查找

class Solution {
    public boolean findNumberIn2DArray(int[][] matrix, int target) {
        if((matrix==null||matrix.length==0)||(matrix.length==1&&matrix[0].length==0))
            return false;
        int i=0,j=matrix[0].length-1;
        while(i<=matrix.length-1&&j>=0){
            if(target==matrix[i][j]) return true;
            if(target>matrix[i][j]) i++;
            else if(target<matrix[i][j]) j--;
        }
        return false;

    } 
}

05、顺时针打印矩阵

class Solution {
    public int[] spiralOrder(int[][] matrix) {
        //判断边界
        if(matrix.length==0)
            return new int[0];//因为返回值是数组，所以实例化一个数组进行返回

        //右下左上
        int left=0,right=matrix[0].length-1,up=0,down=matrix.length-1;
        int num=0;

        int []res=new int[(right+1)*(down+1)];
        while(true){
            //右
            for(int i=left;i<=right;i++)
                res[num++]=matrix[up][i];
            if(++up>down) break;///判断边界

            //下
            for(int i=up;i<=down;i++)
                res[num++]=matrix[i][right];
            if(--right<left) break;
            //左
            for(int i=right;i>=left;i--)
                res[num++]=matrix[down][i];
            if(--down<up) break;

            //上
            for(int i=down;i>=up;i--)
                res[num++]=matrix[i][left];
            if(++left>right) break;
        }
        return res;
    }
}

06、在排序数组中查找数字

• 暴力查找法

class Solution {
    public int search(int[] nums, int target) {
        int res=0;
        for(int i=0;i<nums.length;i++)
                if(nums[i]==target)
                    res++;

        return res;
            
    }
}

• 二分查找法【有序数组优先考虑二分查找】

class Solution {
    public int search(int[] nums, int target) {
       int left = 0, right = nums.length;
        int ans = 0;
        while (left < right){
            int mid = left + ((right - left)/2);
            if (nums[mid] >= target){
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        while (left < nums.length && nums[left++] == target){
            ans++;
        }
        return ans;
            
    }
}

07、剑指 Offer 53 - II. 0～n-1中缺失的数字

class Solution {
    public int missingNumber(int[] nums) {
         int left = 0, right = nums.length;

        while(left < right){
            int mid = left + (right - left) / 2;
            if (nums[mid] == mid) {
                left = mid + 1;
            } else {
                right = mid;
            }  
        }
        return left;
    }
}

二、链表类~

06、从尾到头打印链表

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public int[] reversePrint(ListNode head) {
        ListNode temp=head;
        int size=0;
        while(temp!=null){
            size++;
            temp=temp.next;
        }
        int[] res=new int[size];
        temp=head;
        for(int i=size-1;i>=0;i--){
            res[i]=temp.val;
            temp=temp.next;
        }
        return res;
    }
}

07、删除链表节点

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode deleteNode(ListNode head, int val) {
        if(head.val==val) return head.next;
        ListNode pre=head;
        ListNode cur=head.next;
        while(cur.val!=val&&cur!=null){
            pre=cur;
            cur=cur.next;
        }
        if(cur.val==val){
            pre.next=cur.next;
        }
        return head;
    }
}