Merge pull request #1602 from zhicheng-lee/master

更新 0332.重新安排行程 0051.N皇后 0077.组合优化
This commit is contained in:
程序员Carl
2022-08-31 09:37:36 +08:00
committed by GitHub
3 changed files with 93 additions and 55 deletions

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@ -222,56 +222,6 @@ public:
## 其他语言补充
### Python
```python
class Solution:
def solveNQueens(self, n: int) -> List[List[str]]:
if not n: return []
board = [['.'] * n for _ in range(n)]
res = []
def isVaild(board,row, col):
#判断同一列是否冲突
for i in range(len(board)):
if board[i][col] == 'Q':
return False
# 判断左上角是否冲突
i = row -1
j = col -1
while i>=0 and j>=0:
if board[i][j] == 'Q':
return False
i -= 1
j -= 1
# 判断右上角是否冲突
i = row - 1
j = col + 1
while i>=0 and j < len(board):
if board[i][j] == 'Q':
return False
i -= 1
j += 1
return True
def backtracking(board, row, n):
# 如果走到最后一行,说明已经找到一个解
if row == n:
temp_res = []
for temp in board:
temp_str = "".join(temp)
temp_res.append(temp_str)
res.append(temp_res)
for col in range(n):
if not isVaild(board, row, col):
continue
board[row][col] = 'Q'
backtracking(board, row+1, n)
board[row][col] = '.'
backtracking(board, 0, n)
return res
```
### Java
```java
@ -341,6 +291,55 @@ class Solution {
}
```
### Python
```python
class Solution:
def solveNQueens(self, n: int) -> List[List[str]]:
if not n: return []
board = [['.'] * n for _ in range(n)]
res = []
def isVaild(board,row, col):
#判断同一列是否冲突
for i in range(len(board)):
if board[i][col] == 'Q':
return False
# 判断左上角是否冲突
i = row -1
j = col -1
while i>=0 and j>=0:
if board[i][j] == 'Q':
return False
i -= 1
j -= 1
# 判断右上角是否冲突
i = row - 1
j = col + 1
while i>=0 and j < len(board):
if board[i][j] == 'Q':
return False
i -= 1
j += 1
return True
def backtracking(board, row, n):
# 如果走到最后一行,说明已经找到一个解
if row == n:
temp_res = []
for temp in board:
temp_str = "".join(temp)
temp_res.append(temp_str)
res.append(temp_res)
for col in range(n):
if not isVaild(board, row, col):
continue
board[row][col] = 'Q'
backtracking(board, row+1, n)
board[row][col] = '.'
backtracking(board, 0, n)
return res
```
### Go
```Go
@ -396,6 +395,8 @@ func isValid(n, row, col int, chessboard [][]string) bool {
return true
}
```
### Javascript
```Javascript
var solveNQueens = function(n) {

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@ -20,7 +20,7 @@
大家先回忆一下[77. 组合]给出的回溯法的代码:
```c++
```CPP
class Solution {
private:
vector<vector<int>> result; // 存放符合条件结果的集合
@ -52,7 +52,7 @@ public:
在遍历的过程中有如下代码:
```c++
```CPP
for (int i = startIndex; i <= n; i++) {
path.push_back(i);
backtracking(n, k, i + 1);
@ -76,7 +76,7 @@ for (int i = startIndex; i <= n; i++) {
**如果for循环选择的起始位置之后的元素个数 已经不足 我们需要的元素个数了,那么就没有必要搜索了**
注意代码中i就是for循环里选择的起始位置。
```c++
```CPP
for (int i = startIndex; i <= n; i++) {
```
@ -98,13 +98,13 @@ for (int i = startIndex; i <= n; i++) {
所以优化之后的for循环是
```c++
```CPP
for (int i = startIndex; i <= n - (k - path.size()) + 1; i++) // i为本次搜索的起始位置
```
优化后整体代码如下:
```c++
```CPP
class Solution {
private:
vector<vector<int>> result;

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@ -261,6 +261,43 @@ for (pair<string, int>target : targets[result[result.size() - 1]])
## 其他语言版本
### java
```java
class Solution {
private LinkedList<String> res;
private LinkedList<String> path = new LinkedList<>();
public List<String> findItinerary(List<List<String>> tickets) {
Collections.sort(tickets, (a, b) -> a.get(1).compareTo(b.get(1)));
path.add("JFK");
boolean[] used = new boolean[tickets.size()];
backTracking((ArrayList) tickets, used);
return res;
}
public boolean backTracking(ArrayList<List<String>> tickets, boolean[] used) {
if (path.size() == tickets.size() + 1) {
res = new LinkedList(path);
return true;
}
for (int i = 0; i < tickets.size(); i++) {
if (!used[i] && tickets.get(i).get(0).equals(path.getLast())) {
path.add(tickets.get(i).get(1));
used[i] = true;
if (backTracking(tickets, used)) {
return true;
}
used[i] = false;
path.removeLast();
}
}
return false;
}
}
```
```java
class Solution {