This commit is contained in:
youngyangyang04
2021-08-16 17:27:42 +08:00
9 changed files with 182 additions and 3 deletions

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@ -388,6 +388,44 @@ class Solution:
res += res1
return res
```
动态规划
```python3
class Solution:
def trap(self, height: List[int]) -> int:
leftheight, rightheight = [0]*len(height), [0]*len(height)
leftheight[0]=height[0]
for i in range(1,len(height)):
leftheight[i]=max(leftheight[i-1],height[i])
rightheight[-1]=height[-1]
for i in range(len(height)-2,-1,-1):
rightheight[i]=max(rightheight[i+1],height[i])
result = 0
for i in range(0,len(height)):
summ = min(leftheight[i],rightheight[i])-height[i]
result += summ
return result
```
单调栈
```python3
class Solution:
def trap(self, height: List[int]) -> int:
st =[0]
result = 0
for i in range(1,len(height)):
while st!=[] and height[i]>height[st[-1]]:
midh = height[st[-1]]
st.pop()
if st!=[]:
hright = height[i]
hleft = height[st[-1]]
h = min(hright,hleft)-midh
w = i-st[-1]-1
result+=h*w
st.append(i)
return result
```
Go:

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@ -191,4 +191,57 @@ public:
这里我依然建议大家按部就班把版本一写出来,把情况一二三分析清楚,然后在精简代码到版本二。 直接看版本二容易忽略细节!
## 其他语言版本
Java:
Python:
动态规划
```python3
class Solution:
def largestRectangleArea(self, heights: List[int]) -> int:
result = 0
minleftindex, minrightindex = [0]*len(heights), [0]*len(heights)
minleftindex[0]=-1
for i in range(1,len(heights)):
t = i-1
while t>=0 and heights[t]>=heights[i]: t=minleftindex[t]
minleftindex[i]=t
minrightindex[-1]=len(heights)
for i in range(len(heights)-2,-1,-1):
t=i+1
while t<len(heights) and heights[t]>=heights[i]: t=minrightindex[t]
minrightindex[i]=t
for i in range(0,len(heights)):
left = minleftindex[i]
right = minrightindex[i]
summ = (right-left-1)*heights[i]
result = max(result,summ)
return result
```
单调栈 版本二
```python3
class Solution:
def largestRectangleArea(self, heights: List[int]) -> int:
heights.insert(0,0) # 数组头部加入元素0
heights.append(0) # 数组尾部加入元素0
st = [0]
result = 0
for i in range(1,len(heights)):
while st!=[] and heights[i]<heights[st[-1]]:
midh = heights[st[-1]]
st.pop()
if st!=[]:
minrightindex = i
minleftindex = st[-1]
summ = (minrightindex-minleftindex-1)*midh
result = max(summ,result)
st.append(i)
return result
```
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码.jpg width=450> </img></div>

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@ -159,7 +159,7 @@ public:
return false;
}
que.push(leftNode->left); // 添加p节点的左子树节点
que.push(rightNode->left); // 添加q节点的左子树节点
que.push(rightNode->left); // 添加q节点的左子树节点
que.push(leftNode->right); // 添加p节点的右子树节点
que.push(rightNode->right); // 添加q节点的右子树节点
}

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@ -1205,6 +1205,35 @@ public:
};
```
java代码
```java
class Solution {
public Node connect(Node root) {
Queue<Node> tmpQueue = new LinkedList<Node>();
if (root != null) tmpQueue.add(root);
while (tmpQueue.size() != 0){
int size = tmpQueue.size();
Node cur = tmpQueue.poll();
if (cur.left != null) tmpQueue.add(cur.left);
if (cur.right != null) tmpQueue.add(cur.right);
for (int index = 1; index < size; index++){
Node next = tmpQueue.poll();
if (next.left != null) tmpQueue.add(next.left);
if (next.right != null) tmpQueue.add(next.right);
cur.next = next;
cur = next;
}
}
return root;
}
}
```
python代码

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@ -209,6 +209,22 @@ class Solution(object):
return True
```
Python写法四
```python3
class Solution:
def canConstruct(self, ransomNote: str, magazine: str) -> bool:
c1 = collections.Counter(ransomNote)
c2 = collections.Counter(magazine)
x = c1 - c2
#x只保留值大于0的符号当c1里面的符号个数小于c2时不会被保留
#所以x只保留下了magazine不能表达的
if(len(x)==0):
return True
else:
return False
```
Go
```go

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@ -9,6 +9,7 @@
# 925.长按键入
题目链接https://leetcode-cn.com/problems/long-pressed-name/
你的朋友正在使用键盘输入他的名字 name。偶尔在键入字符 c 按键可能会被长按而字符可能被输入 1 次或多次。
你将会检查键盘输入的字符 typed。如果它对应的可能是你的朋友的名字其中一些字符可能被长按那么就返回 True。
@ -98,7 +99,36 @@ public:
# 其他语言版本
Java
```java
class Solution {
public boolean isLongPressedName(String name, String typed) {
int i = 0, j = 0;
int m = name.length(), n = typed.length();
while (i< m && j < n) {
if (name.charAt(i) == typed.charAt(j)) { // 相同则同时向后匹配
i++; j++;
}
else {
if (j == 0) return false; // 如果是第一位就不相同直接返回false
// 判断边界为n-1,若为n会越界,例如name:"kikcxmvzi" typed:"kiikcxxmmvvzzz"
while (j < n-1 && typed.charAt(j) == typed.charAt(j-1)) j++;
if (name.charAt(i) == typed.charAt(j)) { // j跨越重复项之后再次和name[i]匹配
i++; j++; // 相同则同时向后匹配
}
else return false;
}
}
// 说明name没有匹配完
if (i < m) return false;
// 说明type没有匹配完
while (j < n) {
if (typed.charAt(j) == typed.charAt(j-1)) j++;
else return false;
}
return true;
}
}
```
Python
```python
class Solution:

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@ -169,6 +169,7 @@ class Solution {
}
}
```
Python
```python
class Solution:
def commonChars(self, words: List[str]) -> List[str]:

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@ -30,7 +30,7 @@
2. 空间复杂度是准确算出程序运行时所占用的内存么?
不要以为空间复杂度就已经精准的掌握了程序的内存使用大小,很多因素会影响程序真正内存使用大小,例如编译器的内存对齐,编程语言容器的底层实现等等这些都会影响到程序内存的开销。
不要以为空间复杂度就已经精准的掌握了程序的内存使用大小,很多因素会影响程序真正内存使用大小,例如编译器的内存对齐,编程语言容器的底层实现等等这些都会影响到程序内存的开销。
所以空间复杂度是预先大体评估程序内存使用的大小。

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@ -141,6 +141,18 @@ class Solution:
# 空间复杂度O(n)python的string为不可变需要开辟同样大小的list空间来修改
```
```python 3
#方法三:考虑不能用切片的情况下,利用模+下标实现
class Solution:
def reverseLeftWords(self, s: str, n: int) -> str:
new_s = ''
for i in range(len(s)):
j = (i+n)%len(s)
new_s = new_s + s[j]
return new_s
```
Go
```go