力扣爆刷第122天之CodeTop100五连刷96-100

一、912. 排序数组

题目链接：https://leetcode.cn/problems/sort-an-array/description/
思路：堆排的思想是先从最后一个非叶子节点开始向下递归交换，直到根节点也完成了向下大小交换，此时就构建一个大根堆，只不过左右孩子的顺序不能保证，接下来就从最后一个元素开始与根节点交换，然后向下递归交换，直到所有的节点都完成从上往下交换，即可完成升序排序。

class Solution {public int[] sortArray(int[] nums) {sort(nums);return nums;}// 堆排先构建大顶堆void sort(int[] nums) {// 从第一个非叶子节点开始构建大顶堆for(int i = nums.length/2-1; i >= 0; i--) {adjustSort(nums, i, nums.length);}for(int i = nums.length-1; i > 0; i--) {int t = nums[0];nums[0] = nums[i];nums[i] = t;adjustSort(nums, 0, i);} }// 下降构建大顶堆void adjustSort(int[] nums, int i, int len) {int t = nums[i];for(int k = i*2+1; k < len; k = k*2+1) {if(k + 1 < len && nums[k] < nums[k+1]) {k++;}if(nums[k] > t) {nums[i] = nums[k];i = k;}else{break;}}nums[i] = t;}}

二、24. 两两交换链表中的节点

题目链接：https://leetcode.cn/problems/swap-nodes-in-pairs/description/
思路：用三个指针，前两个负责交换，后一个负责记录下一个位置。

class Solution {public ListNode swapPairs(ListNode head) {if(head == null) return head;ListNode root = new ListNode(-1, head);ListNode a = root, b = root.next, c = b.next;while(c != null) {c = c.next;a.next = b.next;a.next.next = b;b.next = c;if(c != null) {a = b;b = c;c = c.next;}}return root.next;}}

三、297. 二叉树的序列化与反序列化

题目链接：https://leetcode.cn/problems/serialize-and-deserialize-binary-tree/description/
思路：序列化与反序列化，序列化正常前序遍历，需要用占位符划分好节点和null值。反序列化，也是前序遍历构造二叉树，需要解析字符串，方法多样，基于数组也可以做，基于队列也可以。

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
public class Codec {StringBuilder builder = new StringBuilder();LinkedList<String> list = new LinkedList<>();public String serialize(TreeNode root) {traverse(root);return builder.toString();}public TreeNode deserialize(String data) {String[] sList = data.split(",");for(String s : sList) {list.add(s);}return create();}void traverse(TreeNode root) {if(root == null) {builder.append("#,");return;}builder.append(root.val+",");traverse(root.left);traverse(root.right);}TreeNode create() {if(list.size() == 0) {return null;}String s = list.removeFirst();if("#".equals(s)) return null;TreeNode t = new TreeNode(Integer.parseInt(s));t.left = create();t.right = create();return t;}
}// Your Codec object will be instantiated and called as such:
// Codec ser = new Codec();
// Codec deser = new Codec();
// TreeNode ans = deser.deserialize(ser.serialize(root));

四、560. 和为 K 的子数组

题目链接：https://leetcode.cn/problems/subarray-sum-equals-k/description/
思路：用一个全局变量一直记录前缀和，然后用map一直收集前缀和与K的差值，如果差值存在则计数。

class Solution {public int subarraySum(int[] nums, int k) {HashMap<Integer, Integer> map = new HashMap<>();map.put(0, 1);int count = 0, preSum = 0;for(int i = 0; i < nums.length; i++) {preSum += nums[i];if(map.containsKey(preSum - k)) {count += map.get(preSum - k);}map.put(preSum, map.getOrDefault(preSum, 0)+1);}return count;}
}

五、209. 长度最小的子数组

题目链接：https://leetcode.cn/problems/minimum-size-subarray-sum/description/
思路：求最小长度的子数组，很典型的滑动窗口，快慢指针，sum<target时扩大窗口，sum>target时缩小窗口。

class Solution {public int minSubArrayLen(int target, int[] nums) {int slow = 0, fast = 0, sum = 0;int min = Integer.MAX_VALUE;while(fast < nums.length) {sum += nums[fast++];while(sum >= target) {min = Math.min(min, fast - slow);sum -= nums[slow++];}}return min == Integer.MAX_VALUE ? 0 : min;}
}