mirror of
https://github.com/youngyangyang04/leetcode-master.git
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Merge branch 'youngyangyang04:master' into master
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@ -293,85 +293,40 @@ class Solution:
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"""
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Do not return anything, modify board in-place instead.
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"""
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def backtrack(board):
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for i in range(len(board)): #遍历行
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for j in range(len(board[0])): #遍历列
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if board[i][j] != ".": continue
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for k in range(1,10): #(i, j) 这个位置放k是否合适
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if isValid(i,j,k,board):
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board[i][j] = str(k) #放置k
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if backtrack(board): return True #如果找到合适一组立刻返回
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board[i][j] = "." #回溯,撤销k
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return False #9个数都试完了,都不行,那么就返回false
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return True #遍历完没有返回false,说明找到了合适棋盘位置了
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def isValid(row,col,val,board):
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for i in range(9): #判断行里是否重复
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if board[row][i] == str(val):
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self.backtracking(board)
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def backtracking(self, board: List[List[str]]) -> bool:
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# 若有解,返回True;若无解,返回False
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for i in range(len(board)): # 遍历行
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for j in range(len(board[0])): # 遍历列
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# 若空格内已有数字,跳过
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if board[i][j] != '.': continue
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for k in range(1, 10):
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if self.is_valid(i, j, k, board):
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board[i][j] = str(k)
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if self.backtracking(board): return True
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board[i][j] = '.'
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# 若数字1-9都不能成功填入空格,返回False无解
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return False
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return True # 有解
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def is_valid(self, row: int, col: int, val: int, board: List[List[str]]) -> bool:
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# 判断同一行是否冲突
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for i in range(9):
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if board[row][i] == str(val):
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return False
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# 判断同一列是否冲突
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for j in range(9):
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if board[j][col] == str(val):
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return False
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# 判断同一九宫格是否有冲突
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start_row = (row // 3) * 3
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start_col = (col // 3) * 3
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for i in range(start_row, start_row + 3):
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for j in range(start_col, start_col + 3):
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if board[i][j] == str(val):
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return False
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for j in range(9): #判断列里是否重复
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if board[j][col] == str(val):
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return False
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startRow = (row // 3) * 3
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startcol = (col // 3) * 3
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for i in range(startRow,startRow + 3): #判断9方格里是否重复
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for j in range(startcol,startcol + 3):
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if board[i][j] == str(val):
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return False
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return True
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backtrack(board)
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```
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Python3:
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```python3
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class Solution:
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def __init__(self) -> None:
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self.board = []
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def isValid(self, row: int, col: int, target: int) -> bool:
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for idx in range(len(self.board)):
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# 同列是否重复
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if self.board[idx][col] == str(target):
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return False
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# 同行是否重复
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if self.board[row][idx] == str(target):
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return False
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# 9宫格里是否重复
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box_row, box_col = (row // 3) * 3 + idx // 3, (col // 3) * 3 + idx % 3
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if self.board[box_row][box_col] == str(target):
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return False
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return True
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def getPlace(self) -> List[int]:
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for row in range(len(self.board)):
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for col in range(len(self.board)):
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if self.board[row][col] == ".":
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return [row, col]
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return [-1, -1]
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def isSolved(self) -> bool:
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row, col = self.getPlace() # 找个空位置
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if row == -1 and col == -1: # 没有空位置,棋盘被填满的
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return True
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for i in range(1, 10):
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if self.isValid(row, col, i): # 检查这个空位置放i,是否合适
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self.board[row][col] = str(i) # 放i
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if self.isSolved(): # 合适,立刻返回, 填下一个空位置。
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return True
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self.board[row][col] = "." # 不合适,回溯
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return False # 空位置没法解决
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def solveSudoku(self, board: List[List[str]]) -> None:
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"""
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Do not return anything, modify board in-place instead.
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"""
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if board is None or len(board) == 0:
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return
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self.board = board
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self.isSolved()
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```
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Go:
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@ -368,6 +368,41 @@ class Solution:
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return (val1, val2)
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```
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Go:
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> 动态规划
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```go
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func rob(root *TreeNode) int {
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res := robTree(root)
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return max(res[0], res[1])
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func robTree(cur *TreeNode) []int {
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if cur == nil {
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return []int{0, 0}
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}
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// 后序遍历
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left := robTree(cur.Left)
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right := robTree(cur.Right)
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// 考虑去偷当前的屋子
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robCur := cur.Val + left[0] + right[0]
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// 考虑不去偷当前的屋子
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notRobCur := max(left[0], left[1]) + max(right[0], right[1])
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// 注意顺序:0:不偷,1:去偷
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return []int{notRobCur, robCur}
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}
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```
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JavaScript:
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> 动态规划
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@ -226,6 +226,32 @@ class Solution:
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return taraget == dp[taraget]
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```
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Go:
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```go
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// 分割等和子集 动态规划
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// 时间复杂度O(n^2) 空间复杂度O(n)
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func canPartition(nums []int) bool {
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sum := 0
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for _, num := range nums {
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sum += num
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}
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// 如果 nums 的总和为奇数则不可能平分成两个子集
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if sum % 2 == 1 {
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return false
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}
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target := sum / 2
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dp := make([]int, target + 1)
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for _, num := range nums {
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for j := target; j >= num; j-- {
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if dp[j] < dp[j - num] + num {
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dp[j] = dp[j - num] + num
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}
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}
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}
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return dp[target] == target
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}
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```
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```go
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func canPartition(nums []int) bool {
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@ -200,17 +200,14 @@ func reverse(b []byte, left, right int){
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JavaScript:
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```javascript
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var reverseLeftWords = function (s, n) {
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const reverse = (str, left, right) => {
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let strArr = str.split("");
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for (; left < right; left++, right--) {
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[strArr[left], strArr[right]] = [strArr[right], strArr[left]];
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}
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return strArr.join("");
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}
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s = reverse(s, 0, n - 1);
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s = reverse(s, n, s.length - 1);
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return reverse(s, 0, s.length - 1);
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var reverseLeftWords = function(s, n) {
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const length = s.length;
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let i = 0;
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while (i < length - n) {
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s = s[length - 1] + s;
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i++;
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}
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return s.slice(0, length);
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};
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```
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