Merge pull request #1514 from AronJudge/master

0977. 有序数组的平方  0018. 四数之和  0347 前K个高频元素 0104 二叉树最大深度
This commit is contained in:
程序员Carl
2022-07-13 09:38:20 +08:00
committed by GitHub
4 changed files with 20 additions and 18 deletions

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@ -140,6 +140,11 @@ class Solution {
for (int i = 0; i < nums.length; i++) { for (int i = 0; i < nums.length; i++) {
// nums[i] > target 直接返回, 剪枝操作
if (nums[i] > 0 && nums[i] > target) {
return result;
}
if (i > 0 && nums[i - 1] == nums[i]) { if (i > 0 && nums[i - 1] == nums[i]) {
continue; continue;
} }

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@ -294,14 +294,13 @@ class solution {
/** /**
* 递归法 * 递归法
*/ */
public int maxdepth(treenode root) { public int maxDepth(TreeNode root) {
if (root == null) { if (root == null) {
return 0; return 0;
} }
int leftdepth = maxdepth(root.left); int leftDepth = maxDepth(root.left);
int rightdepth = maxdepth(root.right); int rightDepth = maxDepth(root.right);
return math.max(leftdepth, rightdepth) + 1; return Math.max(leftDepth, rightDepth) + 1;
} }
} }
``` ```
@ -311,23 +310,23 @@ class solution {
/** /**
* 迭代法,使用层序遍历 * 迭代法,使用层序遍历
*/ */
public int maxdepth(treenode root) { public int maxDepth(TreeNode root) {
if(root == null) { if(root == null) {
return 0; return 0;
} }
deque<treenode> deque = new linkedlist<>(); Deque<TreeNode> deque = new LinkedList<>();
deque.offer(root); deque.offer(root);
int depth = 0; int depth = 0;
while (!deque.isempty()) { while (!deque.isEmpty()) {
int size = deque.size(); int size = deque.size();
depth++; depth++;
for (int i = 0; i < size; i++) { for (int i = 0; i < size; i++) {
treenode poll = deque.poll(); TreeNode node = deque.poll();
if (poll.left != null) { if (node.left != null) {
deque.offer(poll.left); deque.offer(node.left);
} }
if (poll.right != null) { if (node.right != null) {
deque.offer(poll.right); deque.offer(node.right);
} }
} }
} }

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@ -141,13 +141,10 @@ class Solution {
} }
Set<Map.Entry<Integer, Integer>> entries = map.entrySet(); Set<Map.Entry<Integer, Integer>> entries = map.entrySet();
// 根据map的value值正序排,相当于一个小顶堆 // 根据map的value值构建于一个大顶堆o1 - o2: 小顶堆, o2 - o1 : 大顶堆)
PriorityQueue<Map.Entry<Integer, Integer>> queue = new PriorityQueue<>((o1, o2) -> o1.getValue() - o2.getValue()); PriorityQueue<Map.Entry<Integer, Integer>> queue = new PriorityQueue<>((o1, o2) -> o2.getValue() - o1.getValue());
for (Map.Entry<Integer, Integer> entry : entries) { for (Map.Entry<Integer, Integer> entry : entries) {
queue.offer(entry); queue.offer(entry);
if (queue.size() > k) {
queue.poll();
}
} }
for (int i = k - 1; i >= 0; i--) { for (int i = k - 1; i >= 0; i--) {
result[i] = queue.poll().getKey(); result[i] = queue.poll().getKey();

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@ -106,6 +106,7 @@ class Solution {
int index = result.length - 1; int index = result.length - 1;
while (left <= right) { while (left <= right) {
if (nums[left] * nums[left] > nums[right] * nums[right]) { if (nums[left] * nums[left] > nums[right] * nums[right]) {
// 正数的相对位置是不变的, 需要调整的是负数平方后的相对位置
result[index--] = nums[left] * nums[left]; result[index--] = nums[left] * nums[left];
++left; ++left;
} else { } else {