leetcode 每日刷题 —— 206，169

10.206.反转一个单链表

/**
 * @author mys
 * @version 2019.8.14
 * 反转一个单链表
 */


package com.mys;

import org.junit.Test;

public class ReverseList206 {
    public ListNode reverseList(ListNode head) {
        ListNode prev = null;//前结点
        ListNode curr = head;//当前结点
        while (curr != null) {
            ListNode temp = curr.next;//临时节点，存放当前结点的下一节点

            curr.next = prev;//当前节点 指向前节点
            prev = curr;//前结点向后移
            curr = temp;//当前节向后移
        }
        return prev;
    }

    @Test
    public void fun() {
        //初始化数据
        ListNode a = new ListNode();
        ListNode b = new ListNode();
        ListNode c = new ListNode();
        ListNode d = new ListNode();
        ListNode e = new ListNode();
        a.setVal(10);
        b.setVal(20);
        c.setVal(30);
        d.setVal(40);
        e.setVal(50);
        a.setNext(b);
        b.setNext(c);
        c.setNext(d);
        d.setNext(e);
        e.setNext(null);

        ListNode head = a;
        System.out.println("before reverse:");
        while (head != null) {
            System.out.println(head.val);
            head = head.next;
        }

        System.out.println("after reverse:");
        ListNode reverse = reverseList(a);
        while (reverse != null) {
            System.out.println(reverse.val);
            reverse = reverse.next;
        }
    }
}

class ListNode {
    int val;//存储元素的数据域
    ListNode next;//存储下一个节点的指针域

    public void setVal(int val) {
        this.val = val;
    }

    public void setNext(ListNode next) {
        this.next = next;
    }

    public ListNode() {

    }

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

11.169.给定一个大小为 n 的数组，找到其中的众数。众数是指在数组中出现次数大于 ⌊ n/2 ⌋ 的元素。
你可以假设数组是非空的，并且给定的数组总是存在众数

public class MajorityElement169 {
    //摩尔投票法
    //假设第一个数过半数，后面如果相同加一，不同减一
    //当count减为0时，重新更换数
    public int majorityElement(int[] nums) {
        int res = 0;
        int count = 0;

        for (int i = 0; i < nums.length; i ++) {
            if (count == 0) {
                res = nums[i];
                count ++;
            }else {
                if (res == nums[i]) {
                    count++;
                } else {
                    count--;
                }
            }
        }

        return res;
    }