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LeetCode 203:根据值删除节点

题目:
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方式一:
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方法二:
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代码示例:

package com.zy.leetcode.LeetCode_203;

/**
 * @Author: zy
 * @Date: 2024-12-25-11:49
 * @Description:
 * 根据值删除节点
 */
public class ListNode_203 {

    int val;
    ListNode_203 next;

    ListNode_203(int val) {
        this.val = val;
        this.next = null;
    }

    public ListNode_203(int val, ListNode_203 next) {
        this.val = val;
        this.next = next;
    }

    public ListNode_203() {

    }

    public ListNode_203 removeVal(ListNode_203 o1, int target) {
        ListNode_203 s = new ListNode_203(-1, o1);
        ListNode_203 p1 = s;
        ListNode_203 p2 = s.next;

        while (p2 != null) {
            if (p2.val == target) {
                //删除,p2向后平移
                p1.next = p2.next;
            } else {
                p1 = p1.next;
            }
            p2 = p2.next;
        }
        //返回头节点,不返回
        return s.next;

    }

    public ListNode_203 removeValChange(ListNode_203 o1, int target) {
        ListNode_203 s = new ListNode_203(-1, o1);
        ListNode_203 p1 = s;
        ListNode_203 p2;

        while ((p2 = p1.next) != null) {
            if (p2.val == target) {
                //删除,p2向后平移
                p1.next = p2.next;
            } else {
                p1 = p1.next;
            }
            //p2 = p2.next;
        }
        //返回头节点,不返回
        return s.next;

    }

    /**
     * 递归
     * @param o1
     * @param target
     * @return
     */
    public ListNode_203 removeVal2(ListNode_203 o1, int target) {
        if (o1 == null) {
            return null;
        }

        if (o1.val == target) {
            //如果相等,则返回后续节点递归结果
            return removeVal2(o1.next, target);
        } else {
            //如果不相等,则返回后续节点,并更新当前节点所指向的下一节点
            o1.next = removeVal2(o1.next, target);
            return o1;
        }

    }

    // Helper method to print the linked list
    private static void printLinkedList(ListNode_203 head) {
        ListNode_203 current = head;
        while (current != null) {
            System.out.print(current.val + " ");
            current = current.next;
        }
        System.out.println(); // For a new line after printing the list
    }

    // Helper method to create a linked list from an array of values
    private static ListNode_203 createLinkedList(int[] values) {
        if (values.length == 0) {
            return null;
        }

        ListNode_203 head = new ListNode_203(values[0]);
        ListNode_203 current = head;

        for (int i = 1; i < values.length; i++) {
            current.next = new ListNode_203(values[i]);
            current = current.next;
        }

        return head;
    }

    public static void main(String[] args) {
        int[] values1 = {2, 2, 3, 4, 2, 5, 6};
        int[] values2 = {1, 2, 6, 3, 6};
        int[] values3 = {7, 7, 7, 7, 7};
        ListNode_203 head1 = createLinkedList(values1);
        printLinkedList(head1);

        ListNode_203 head2 = createLinkedList(values2);
        printLinkedList(head2);

        ListNode_203 head3 = createLinkedList(values3);
        printLinkedList(head3);

        System.out.println("\n-------------");
        ListNode_203 listNode1 = new ListNode_203().removeVal(head1, 2);
        ListNode_203 listNode12 = new ListNode_203().removeValChange(head1, 2);
        ListNode_203 listNode13 = new ListNode_203().removeVal2(head1, 2);
        printLinkedList(listNode1);
        printLinkedList(listNode12);
        printLinkedList(listNode13);

        System.out.println("\n-------------");
        ListNode_203 listNode2 = new ListNode_203().removeVal(head2, 6);
        printLinkedList(listNode2);

        System.out.println("\n-------------");
        ListNode_203 listNode3 = new ListNode_203().removeVal(head3, 7);
        printLinkedList(listNode3);

    }
}


原文地址:https://blog.csdn.net/qq_41041630/article/details/144803742

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