【华为OJ】【093-从单向链表中删除指定值的节点】

【华为OJ】【算法总篇章】

【华为OJ】【093-从单向链表中删除指定值的节点】

【工程下载】

题目描述

输入一个单向链表和一个节点的值，从单向链表中删除等于该值的节点，删除后如果链表中无节点则返回空指针。
链表结点定义如下：
struct ListNode
{
int       m_nKey;
ListNode* m_pNext;
};
详细描述：
本题为考察链表的插入和删除知识。
链表的值不能重复
构造过程，例如
1 -> 2
3 -> 2
5 -> 1
4 -> 5
7 -> 2
最后的链表的顺序为 2 7 3 1 5 4
删除 结点 2
则结果为 7 3 1 5 4

输入描述:

1 输入链表结点个数
2 输入头结点的值
3 按照格式插入各个结点
4 输入要删除的结点的值

输出描述:

输出删除结点后的序列

输入例子:

5
2
3 2
4 3
5 2
1 4
3

输出例子:

2 1 5 4

算法实现

import java.util.List;
import java.util.Scanner;

/**
* Author: 王俊超
* Date: 2016-05-03 09:47
* CSDN: http://blog.csdn.net/derrantcm * Github: https://github.com/Wang-Jun-Chao * Declaration: All Rights Reserved !!!
*/
public class Main {
private static class ListNode {
int key;
ListNode next;
}

public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
//        Scanner scanner = new Scanner(Main.class.getClassLoader().getResourceAsStream("data.txt"));
while (scanner.hasNextLong()) {
int num = scanner.nextInt();
int h = scanner.nextInt();
ListNode head = new ListNode();
head.key = h;

for (int i = 0; i < num - 1; i++) {
int newVal = scanner.nextInt();
int afterVal = scanner.nextInt();
addNode(newVal, afterVal, head);
}

int del = scanner.nextInt();

head = delete(del, head);
System.out.println(getString(head));

}
scanner.close();
}

private static String getString(ListNode head) {
StringBuilder builder = new StringBuilder();

while (head!= null) {
builder.append(head.key).append(' ');
head = head.next;
}

//        return builder.substring(0, builder.length() - 1);
return builder.toString();
}

private static void addNode(int newVal, int afterVal, ListNode head) {
ListNode n = head;
while (n != null) {
if (n.key == afterVal) {
ListNode node = new ListNode();
node.key = newVal;
node.next = n.next;
n.next = node;
break;
}
n = n.next;
}
}

private static ListNode delete(int val, ListNode head) {
if (head.key == val) {
ListNode ret = head.next;
head.next = null;
return ret;
} else {
ListNode prev = head;
while (prev.next != null) {
if (prev.next.key == val) {
prev.next = prev.next.next;
break;
}
prev = prev.next;
}

return head;
}
}
}