LeetCode -- Flora -- edit 2025-04-17
1.最长连续序列
128. 最长连续序列
给定一个未排序的整数数组 nums ，找出数字连续的最长序列（不要求序列元素在原数组中连续）的长度。
请你设计并实现时间复杂度为 O(n) 的算法解决此问题。
示例 1：
输入：nums = [100,4,200,1,3,2] 输出：4 解释：最长数字连续序列是 [1, 2, 3, 4]。它的长度为 4。
示例 2：
输入：nums = [0,3,7,2,5,8,4,6,0,1] 输出：9
示例 3：
输入：nums = [1,0,1,2] 输出：3
提示：
0 <= nums.length <= 105-109 <= nums[i] <= 109
public static void main(String[] args) {int[] nums1 = {100,4,200,1,3,2};//排序int[] nums2 = {0};//1个数int[] nums3 = {0,-1};//绝对值==1 连续int[] nums = {0,3,7,2,5,8,4,6,0,1};//去重int longestConsecutive = longestConsecutive2(nums);System.out.println(longestConsecutive);}//1-answerpublic static int longestConsecutive2(int[] nums) {if(nums.length == 0){return 0;}Arrays.sort(nums);//排序int longestStreak = 1;//最大连续长度int currentStreak = 1;//当前连续长度for(int i = 1; i < nums.length; i++){if(nums[i] != nums[i-1]){//去重if(nums[i] == nums[i-1] + 1){//差值1，连续currentStreak += 1;}else{longestStreak = Math.max(longestStreak, currentStreak);//判断更新currentStreak = 1;//重置}}}return Math.max(longestStreak, currentStreak);}//1-mypublic static int longestConsecutive(int[] nums) {Arrays.sort(nums);int longest = 0;List<List<Integer>> enterList = new ArrayList<>();//起始，长度for (int i = 0;i<nums.length; i++){List<Integer> list = new ArrayList<>();list.add(nums[i]);int left = 0;int right = 0;for (int j = i+1; j<nums.length ;j++){left = nums[j-1];right = nums[j];if (left==right){continue;}if ( Math.abs(left-right) == 1){list.add(nums[j]);}if (Math.abs(left-right) != 1){break;}}enterList.add(list);i = i+list.size()-1;}for (List<Integer> o : enterList) {//空间利用率很差if (o.size()>longest){longest = o.size();}}return longest;}2.移动零
283. 移动零
给定一个数组 nums，编写一个函数将所有 0 移动到数组的末尾，同时保持非零元素的相对顺序。
请注意 ，必须在不复制数组的情况下原地对数组进行操作。
示例 1:
输入: nums =[0,1,0,3,12]输出:[1,3,12,0,0]
示例 2:
输入: nums =[0]输出:[0]
提示:
1 <= nums.length <= 104-231 <= nums[i] <= 231 - 1
进阶：你能尽量减少完成的操作次数吗？
public static void main(String[] args) {int[] nums = {0,3,7,2,5,8,4,6,0,1};//去重moveZeroes2(nums);System.out.println(Arrays.toString(nums));}//2-answerpublic static void moveZeroes2(int[] nums) {int i=0,j=0;while(j<nums.length){if(nums[j] != 0){nums[i] = nums[j];i++;//indexAfterMove}j++;//index}while(i < nums.length){nums[i++] = 0;}}//2-mypublic static void moveZeroes(int[] nums) {int zeroCount = 0;Map<Integer,Integer> rightIndexAndNumMap = new HashMap<>();for (int i = 0; i < nums.length; i++) {if (nums[i] == 0){zeroCount++;}if (nums[i]!=0 && zeroCount>0){//开始移动rightIndexAndNumMap.put(i-zeroCount,nums[i]);}}for (Map.Entry<Integer, Integer> entry : rightIndexAndNumMap.entrySet()) {nums[entry.getKey()] = entry.getValue();}for (int i = 0; i < zeroCount; i++) {nums[nums.length-i-1] = 0;}}