5. Numbers that only appear once
Given an array of non-empty integers, except for an element that appears only once, every other element appears twice. Find the element that appears only once.
Explanation:
Your algorithm should have linear time complexity. Can you do it without using extra space?
Method 1: Mine, but using sorting, the complexity should not meet the linear requirements
class Solution {
public int singleNumber(int[] nums) {
if(nums.length<2){
return nums[0];
}
Arrays.sort(nums);
if(nums[0]!=nums[1]){
return nums[0];
}
for(int i=1;i<nums.length-1;i++){
if(nums[i]!=nums[i-1] && nums[i]!=nums[i+1]){
return nums[i];
}
}
return nums[nums.length-1];
}
}
Method 2: XOR operation
class Solution {
public int singleNumber(int[] nums) {
int single = 0;
for (int num : nums) {
single = single ^ num;
}
return single;
}
}
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/single-number/solution/zhi-chu-xian-yi-ci-de-shu-zi-by-leetcode-solution/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
Method 3: Hash set
public static int SingleNumber(int[] nums)
{
HashSet<int> set = new HashSet<int>();
for (int i = 0; i < nums.Length; i++)
{
//该判断语句的整体作用是：如果当前数字（nums[i]）已经在之前出现过，那么在哈希集实例（set）中移除当前数字
// Add 方法的作用是添加当前数字于哈希集中，如果当前数字和该集合（set）元素存在重复，则返回 False 。故在此采用了逻辑非操作符（!）
if (!set.Add(nums[i]))
set.Remove(nums[i]); ;
}
//因为每个重复元素最多存在两个，而重复元素的第一个添加后均被移除，而第二个均未添加成功，故此时哈希集只保留唯一且未重复的元素
// First*1 方法的作用是返回该序列的第一个元素
return set.First();
}
作者：magicalchao
链接：https://leetcode-cn.com/problems/single-number/solution/cou-yi-pian-ti-jie-hua-shuo-ti-jie-hen-hao-wan-by-/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
Six, the intersection of two arrays II
Given two arrays, write a function to calculate their intersection.
Method 1: Official, hash table. Use a hash table to store the number of occurrences of each number.
class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
//把小的数组放前面
if (nums1.length > nums2.length) {
return intersect(nums2, nums1);
}
//创建一个哈希表
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
//如果表中有了，就把值取出来加1；如果表中没有，就0加1
for (int num : nums1) {
int count = map.getOrDefault(num, 0) + 1;
map.put(num, count);
}
int[] intersection = new int[nums1.length];
int index = 0;
//把num2中的值一一取出来和哈希表中比较
for (int num : nums2) {
int count = map.getOrDefault(num, 0);
//如果count>0，则代表哈希表中有这个数
if (count > 0) {
intersection[index++] = num;
count--;
if (count > 0) {
map.put(num, count);
} else {
map.remove(num);
}
}
}
return Arrays.copyOfRange(intersection, 0, index);
}
}
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/intersection-of-two-arrays-ii/solution/liang-ge-shu-zu-de-jiao-ji-ii-by-leetcode-solution/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
Method 2: Aiming at the above method, I reproduced and improved it
class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
if(nums1.length>nums2.length){
return intersect(nums2,nums1);
}
Map<Integer,Integer> map=new HashMap<Integer,Integer>();
for(int num:nums1){
int count=map.getOrDefault(num, 0) + 1;
map.put(num,count);
}
int[] willback = new int[nums1.length];
int index=0;
for(int num:nums2){
int count = map.getOrDefault(num, 0);
if(count>0){
willback[index]=num;
index++;
map.put(num,count-1);
}
}
return Arrays.copyOfRange(willback, 0, index);
}
}
Method 3: Use double pointers after sorting
class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
Arrays.sort(nums1);
Arrays.sort(nums2);
int length1=nums1.length;
int length2=nums2.length;
int[] willback = new int[Math.min(length1,length2)];
int index1=0;
int index2=0;
int index3=0;
while (index1 < length1 && index2 < length2) {
if(nums1[index1]<nums2[index2]){
index1++;
}
else if(nums1[index1]>nums2[index2]){
index2++;
}
else{
willback[index3]=nums1[index1];
index3++;
index1++;
index2++;
}
}
return Arrays.copyOfRange(willback,0,index3);
}
}