一、字符串处理

题目描述

Given an integer, convert it to a roman numeral.
Input is guaranteed to be within the range from 1 to 3999.

思路

把数字转化为罗马符号，根据罗马符号的规律，可以先用map来存储一下。之后把每一位添加到所求中去。
语法点：StringBuffer是字符缓冲区，是可以修改字符长度的，最后要用sb.toString()去返回缓冲区的字符串

代码

//！COPY

public class Solution {
    public String intToRoman(int num) {
             String[][] map={
            {" ","I ","II ","III ","IV ","V ","VI ","VII ","VIII ","IX "},
            {" ","X ","XX ","XXX ","XL ","L ","LX ","LXX ","LXXX ","XC "},
            {" ","C ","CC ","CCC ","CD ","D ","DC ","DCC ","DCCC ","CM "},
            {" ","M ","MM ","MMM "}
        };
        StringBuffer sb=new StringBuffer();
        sb.append(map[3][num/1000%10]);
        sb.append(map[2][num/100%10]);
        sb.append(map[1][num/10%10]);
        sb.append(map[0][num%10]);
        
        return sb.toString();
    }
}

二、链表

题目描述

You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

思路

链表的题，虽然题目说存放的数字是反过来的，但是完全可以把加法也”反过来“，也就是说从”左到右“相加。
链表的题，一般都应该设置一个头指针head（里面什么也不存放，只是用来记住链表的头）

代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if(l1==null){
            return l2;
        }
        if(l2==null){
            return l1;
        }
        ListNode head=new ListNode(0);
        ListNode p=head;
        int tem=0;
        while(l1!=null||l2!=null||tem!=0){
            if(l1!=null){
                tem+=l1.val;
                l1=l1.next;
            }
            if(l2!=null){
                tem+=l2.val;
                l2=l2.next;
            }
            p.next=new ListNode(tem%10);
            p=p.next;
            tem/=10;
        }
        return head.next;
    }
}

三、最长无重复字符子串

题目描述

Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb " is "abc ", which the length is 3. For "bbbbb " the longest substring is "b ", with the length of 1.

思路

水题渐渐做完了，开始碰到的题有难度了。这题用到了所谓的滑动窗口法：从左到右滑动，如果碰到了在左边界内且出现过的字符，那就将左边界移动到之前的那个字符的下一位，刷新求最大即可。
还看到一位大神的神解法，代码很简洁，思想其实也是一样的：从左到右滑动，记录每一个字符上一次出现的位置，在第i位时比较当前字符的上一次出现的位置和左边界，刷新当前左边界。

代码1

import java.util.HashMap;

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        HashMap <Character,Integer> map=new HashMap<>();
        int len=s.length();
        if(s==null||len==0){
            return 0;
        }
        int res=0;
        int l=0;
        for(int i=0;i<len;i++){
            char c=s.charAt(i);
            if(map.containsKey(c)){
                l=Math.max(l,map.get(c)+1);
            }
            res=Math.max(res,i-l+1);
            map.put(c,i);
        }
        return res;
    }
}

代码2

import java.util.HashMap;

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        HashMap<Character,Integer> map=new HashMap<>();
        int len = s.length();
        if(s==null||len==0){
            return 0;
        }
        for(int i=0;i<len;i++){
            map.put(s.charAt(i),-1);
        }
        int l=-1;
        int res=0;
        for(int i=0;i<len;i++){
            char c=s.charAt(i);
            l=Math.max(l,map.get(c));
            res=Math.max(res,i-l);
            map.put(c,i);
        }
        return res;
    }
}

leetcode算法题解(Java版)-6-链表，字符串

一、字符串处理

二、链表

三、最长无重复字符子串