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Commit 6df2438

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author
luzhipeng
committed
2 parents f1c88c5 + 03ebc74 commit 6df2438

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3 files changed

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-37
lines changed

3 files changed

+44
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‎problems/104.maximum-depth-of-binary-tree.md‎

Lines changed: 18 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -49,7 +49,7 @@ var maxDepth = function(root) {
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- 树的基本操作- 遍历 - 层次遍历(BFS)
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## 代码
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* 语言支持:JS,C++
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* 语言支持:JS,C++,Python
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JavaScript Code:
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```js
@@ -130,6 +130,23 @@ public:
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}
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};
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```
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Python Code:
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```python
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class Solution:
137+
def maxDepth(self, root: TreeNode) -> int:
138+
if not root: return 0
139+
q, depth = [root, None], 1
140+
while q:
141+
node = q.pop(0)
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if node:
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if node.left: q.append(node.left)
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if node.right: q.append(node.right)
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elif q:
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q.append(None)
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depth += 1
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return depth
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```
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## 相关题目
134151
- [102.binary-tree-level-order-traversal](./102.binary-tree-level-order-traversal.md)
135152
- [103.binary-tree-zigzag-level-order-traversal](./103.binary-tree-zigzag-level-order-traversal.md)

‎problems/121.best-time-to-buy-and-sell-stock.md‎

Lines changed: 23 additions & 33 deletions
Original file line numberDiff line numberDiff line change
@@ -40,7 +40,7 @@ Explanation: In this case, no transaction is done, i.e. max profit = 0.
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## 代码
4242

43-
语言支持:JS,Python,C++
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语言支持:JS,Python,C++
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JS Code:
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@@ -60,31 +60,31 @@ JS Code:
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*
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* Say you have an array for which the i^th element is the price of a given
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* stock on day i.
63-
*
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*
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* If you were only permitted to complete at most one transaction (i.e., buy
6565
* one and sell one share of the stock), design an algorithm to find the
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* maximum profit.
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*
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*
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* Note that you cannot sell a stock before you buy one.
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*
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*
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* Example 1:
71-
*
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*
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*
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*
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* Input: [7,1,5,3,6,4]
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* Output: 5
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* Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit
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* = 6-1 = 5.
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* Not 7-1 = 6, as selling price needs to be larger than buying price.
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*
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*
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*
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*
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* Example 2:
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*
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*
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*
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*
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* Input: [7,6,4,3,1]
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* Output: 0
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* Explanation: In this case, no transaction is done, i.e. max profit = 0.
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*
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*
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*
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*
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*/
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/**
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* @param {number[]} prices
@@ -111,28 +111,19 @@ var maxProfit = function(prices) {
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Python Code:
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```python
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# 应用Kadane's algorithms
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class Solution:
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def maxProfit(self, prices: 'List[int]') -> int:
117-
"""
118-
step by step
119-
"""
120-
# error case
121-
if len(prices) < 1:
122-
return 0
123-
124-
# caluate the daily gains, break into a subarray problem
125-
gains = [prices[i]-prices[i-1] for i in range(1, len(prices))]
126-
127-
loc_max = global_max = 0 #not gains[0] in case of negative
128-
for i in range(len(gains)):
129-
loc_max = max(loc_max + gains[i], gains[i])
130-
if loc_max > global_max:
131-
global_max = loc_max
132-
"""
133-
Runtime: 48 ms, faster than 34.50% of Python3 online submissions for Best Time to Buy and Sell Stock.
134-
Memory Usage: 14.1 MB, less than 10.26% of Python3 online submissions for Best Time to Buy and Sell Stock.
135-
"""
116+
if not prices: return 0
117+
118+
min_price = float('inf')
119+
max_profit = 0
120+
121+
for price in prices:
122+
if price < min_price:
123+
min_price = price
124+
elif max_profit < price - min_price:
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max_profit = price - min_price
126+
return max_profit
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```
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C++ Code:
@@ -164,4 +155,3 @@ public:
164155
165156
- [122.best-time-to-buy-and-sell-stock-ii](./122.best-time-to-buy-and-sell-stock-ii.md)
166157
- [309.best-time-to-buy-and-sell-stock-with-cooldown](./309.best-time-to-buy-and-sell-stock-with-cooldown.md)
167-

‎problems/88.merge-sorted-array.md‎

Lines changed: 3 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -147,9 +147,9 @@ function merge(nums1, nums2) {
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* @return {void} Do not return anything, modify nums1 in-place instead.
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*/
149149
var merge = function(nums1, m, nums2, n) {
150-
// 设置一个指针,指针初始化指向nums1的末尾
150+
// 设置一个指针,指针初始化指向nums1的末尾(根据#62,应该是index为 m+n-1 的位置,因为nums1的长度有可能更长)
151151
// 然后不断左移指针更新元素
152-
let current = nums1.length - 1;
152+
let current = m + n - 1;
153153

154154
while (current >= 0) {
155155
// 没必要继续了
@@ -181,7 +181,7 @@ C++ code:
181181
class Solution {
182182
public:
183183
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
184-
int current = nums1.size()- 1;
184+
int current = m + n - 1;
185185
while (current >= 0) {
186186
if (n == 0) return;
187187
if (m < 1) {

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