1.字符串中第一个只出现一次字符

class Solution {public int firstUniqChar(String s) {int[] count=new int[26];for(int i=0;i<s.length();i++){char ch=s.charAt(i);count[ch-'a']++;}for(int i=0;i<s.length();i++){char ch=s.charAt(i);if(count[ch-'a']==1){return i;}}return -1;}  
}

387. 字符串中的第一个唯一字符 - 力扣（LeetCode）

2. 只出现一次的数字

136. 只出现一次的数字 - 力扣（LeetCode）

class Solution {public int singleNumber(int[] nums) {Set<Integer> set=new HashSet<>();for(int i=0;i<nums.length;i++){if(set.contains(nums[i])){set.remove(nums[i]);}else{set.add(nums[i]);}}for(int i=0;i<nums.length;i++){if(set.contains(nums[i])){return nums[i];}}return -1;}
}

 3.随机链表的复制

138. 随机链表的复制 - 力扣（LeetCode）

/*
// Definition for a Node.
class Node {int val;Node next;Node random;public Node(int val) {this.val = val;this.next = null;this.random = null;}
}
*/class Solution {public Node copyRandomList(Node head) {HashMap<Node,Node> map=new HashMap<>();Node cur=head;while(cur !=null){Node node=new Node(cur.val);map.put(cur,node);cur=cur.next;}cur=head;while(cur!=null){map.get(cur).next=map.get(cur.next);map.get(cur).random=map.get(cur.random);cur=cur.next;}return map.get(head);}
}

 4.宝石与石头

771. 宝石与石头 - 力扣（LeetCode）

class Solution {public int numJewelsInStones(String jewels, String stones) {Set<Character> set=new HashSet<>();for(char ch:jewels.toCharArray()){set.add(ch);}int count=0;for(char ch:stones.toCharArray()){if(set.contains(ch)){count++;}}return count;}
}

5.坏键盘

旧键盘 (20)__牛客网 (nowcoder.com)

import java.util.*;
// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {public static void main(String[] args) {Scanner in = new Scanner(System.in);// 注意 hasNext 和 hasNextLine 的区别while (in.hasNextLine()) { // 注意 while 处理多个 caseString s1=in.nextLine();String s2=in.nextLine();func(s1,s2);}}public static void func(String s1,String s2){Set<Character> set=new HashSet<>();for(char ch:s2.toUpperCase().toCharArray()){set.add(ch);}Set<Character> set1=new HashSet<>();for(char ch:s1.toUpperCase().toCharArray()){if(!set.contains(ch)&&!set1.contains(ch)){System.out.print(ch);set1.add(ch);}}}
}

6.前k个高频单词

692. 前K个高频单词 - 力扣（LeetCode）

1.统计每个单词出现的次数

2.放入优先级队列中（建立小根堆）

public List<String> topKFrequent(String[] words, int k) {//1、统计每个单词出现的次数Map<String,Integer> map = new HashMap<>();for (String word : words) {if(map.get(word) == null) {map.put(word,1);}else {int val = map.get(word);map.put(word,val+1);}}//2、建立小根堆,指定比较的方式PriorityQueue<Map.Entry<String,Integer>> minHeap = new PriorityQueue<>(new Comparator<Map.Entry<String, Integer>>() {@Overridepublic int compare(Map.Entry<String, Integer> o1, Map.Entry<String, Integer> o2) {if(o1.getValue().compareTo(o2.getValue()) == 0) {//按照字母顺序建立大根堆return o2.getKey().compareTo(o1.getKey());}return o1.getValue()-o2.getValue();}});//3、遍历map 调整优先级队列for(Map.Entry<String,Integer> entry: map.entrySet()) {if(minHeap.size() < k) {minHeap.offer(entry);}else {Map.Entry<String,Integer> top = minHeap.peek();//如果当前频率相同if(top.getValue().compareTo(entry.getValue()) == 0) {// 字母顺序小的进来if(top.getKey().compareTo(entry.getKey()) > 0) {//出队minHeap.poll();minHeap.offer(entry);}}else {if(top.getValue().compareTo(entry.getValue()) < 0) {minHeap.poll();minHeap.offer(entry);}}}}List<String> ret = new ArrayList<>();for (int i = 0; i < k; i++) {Map.Entry<String,Integer> top  = minHeap.poll();ret.add(top.getKey());}Collections.reverse(ret);return ret;}