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Merged

pull merged 6 commits into AlgorithmAndLeetCode:master from youngyangyang04:master

May 15, 2023

Merged

[pull] master from youngyangyang04:master #265

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Update 0225.用队列实现栈.md

Lozakaka Mar 30, 2023

修复 0101.对称二叉树.md 里的错别字

enjoy-binbin Apr 25, 2023

fix 0343 py code

xiaodi007 Apr 28, 2023

Merge pull request #2039 from Lozakaka/patch-1

youngyangyang04 May 15, 2023

Merge pull request #2057 from enjoy-binbin/fix_typo

youngyangyang04 May 15, 2023

Merge pull request #2060 from xiaodi007/fix_0343_py

youngyangyang04 May 15, 2023

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4 changes: 2 additions & 2 deletions problems/0101.对称二叉树.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -88,7 +88,7 @@ else if (left->val != right->val) return false; // 注意这里我没有


		* 比较二叉树外侧是否对称:传入的是左节点的左孩子,右节点的右孩子。
	* 比较内测是否对称,传入左节点的右孩子,右节点的左孩子。
	* 比较内侧是否对称,传入左节点的右孩子,右节点的左孩子。
		* 如果左右都对称就返回true ,有一侧不对称就返回false 。

		代码如下:
Expand Down Expand Up		@@ -157,7 +157,7 @@ public:

		这个代码就很简洁了,但隐藏了很多逻辑,条理不清晰,而且递归三部曲,在这里完全体现不出来。

	所以建议大家做题的时候,一定要想清楚逻辑,每一步做什么。把道题目所有情况想到位,相应的代码写出来之后,再去追求简洁代码的效果。
	所以建议大家做题的时候,一定要想清楚逻辑,每一步做什么。把题目所有情况想到位,相应的代码写出来之后,再去追求简洁代码的效果。

		## 迭代法

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37 changes: 37 additions & 0 deletions problems/0225.用队列实现栈.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -365,6 +365,43 @@ class MyStack {
		}
		}

	```
	优化,使用一个 Queue 实现,但用卡哥的逻辑实现
	```
	class MyStack {
	Queue<Integer> queue;

	public MyStack() {
	queue = new LinkedList<>();
	}

	public void push(int x) {
	queue.add(x);
	}

	public int pop() {
	rePosition();
	return queue.poll();
	}

	public int top() {
	rePosition();
	int result = queue.poll();
	queue.add(result);
	return result;
	}

	public boolean empty() {
	return queue.isEmpty();
	}

	public void rePosition(){
	int size = queue.size();
	size--;
	while(size-->0)
	queue.add(queue.poll());
	}
	}
		```

		Python:
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2 changes: 1 addition & 1 deletion problems/0343.整数拆分.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -254,7 +254,7 @@ class Solution:
		# 假设对正整数 i 拆分出的第一个正整数是 j(1 <= j < i),则有以下两种方案:
		# 1) 将 i 拆分成 j 和 i−j 的和,且 i−j 不再拆分成多个正整数,此时的乘积是 j * (i-j)
		# 2) 将 i 拆分成 j 和 i−j 的和,且 i−j 继续拆分成多个正整数,此时的乘积是 j * dp[i-j]
	for j in range(1, i / 2 + 1):
	for j in range(1, i // 2 + 1):
		dp[i] = max(dp[i], max(j * (i - j), j * dp[i - j]))
		return dp[n]
		```
Expand Down