LeetCode经典编程题（五）

1. copy-list-with-random-pointer

Description

A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.

Return a deep copy of the list.

Solution

Create all nodes then link the random and next node.

Code

/**
 * Definition for singly-linked list with a random pointer.
 * class RandomListNode {
 *     int label;
 *     RandomListNode next, random;
 *     RandomListNode(int x) { this.label = x; }
 * };
 */
import java.util.ArrayList;
public class Solution {
    public RandomListNode copyRandomList(RandomListNode head) {
        //create all nodes
        ArrayList<RandomListNode> list = new ArrayList<>();
        ArrayList<Integer> labels = new ArrayList<>();
        RandomListNode p = head;
        while(p!=null){
            list.add(new RandomListNode(p.label));
            labels.add(p.label);
            p = p.next;
        }
        list.add(null);
        
        p = head;
        int pointer = 0;
        while(p!=null){
            if(p.random!=null)
                list.get(pointer).random = list.get(labels.indexOf(p.random.label));
            else
                list.get(pointer).random = null;
            list.get(pointer).next = list.get(pointer+1);
            pointer++;
            p = p.next;
        }
        return list.get(0);
    }
}

2. single-number-ii

Description

Given an array of integers, every element appears three times except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Solution

reference: https://www.nowcoder.com/profile/2343147/codeBookDetail?submissionId=20452002

Code

without using extra memory

public int singleNumber(int[] A) {
	int ones = 0, twos = 0;
	for(int i = 0; i < A.length; i++){
		ones = (ones ^ A[i]) & ~twos;
		twos = (twos ^ A[i]) & ~ones;
	}
	return ones;
}

use extra memory

import java.util.*;
public class Solution {
    public int singleNumber(int[] A) {
        Map<Integer, Boolean> map = new HashMap<>();
        for(int i = 0; i < A.length; i++){
            if(map.containsKey(A[i])){
                map.put(A[i], false);
            }else{
                map.put(A[i], true);
            }
        }
        for(Integer key: map.keySet()){
            if(map.get(key)){
                return key;
            }
        }
        return -1;
    }
}

3. single-number

Description

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Code

public class Solution {
    public int singleNumber(int[] A) {
        int num = 0;
        for(int i=0;i<A.length;i++){
            num = (num ^ A[i]);
        }
        return num;
    }
}