[Leetcode][python]Spiral Matrix/Spiral Matrix II/螺旋矩阵/螺旋矩阵 II

旧城等待， 2022-06-07 03:55 160阅读 0赞

# Spiral Matrix #

## 题目大意 ##

将一个矩阵中的内容螺旋输出。  
注意点：

矩阵不一定是正方形  
例子：

输入: matrix = \[ \[1, 2, 3\], \[4, 5, 6\], \[7, 8, 9\] \]  
输出: \[1, 2, 3, 6, 9, 8, 7, 4, 5\]

## 解题思路 ##

看代码就可以理解  
注意：

while True：  
然后里面自行判断结束

## 代码 ##

class Solution(object):
        def spiralOrder(self, matrix):
            """
            :type matrix: List[List[int]]
            :rtype: List[int]
            """
            if not matrix: 
                return []
            up = 0
            left = 0
            down = len(matrix)-1
            right = len(matrix[0])-1
            direct = 0  # 0: go right   1: go down  2: go left  3: go up
            res = []
            while True:
                if direct == 0:
                    for i in range(left, right+1):
                        res.append(matrix[up][i])
                    up += 1
                if direct == 1:
                    for i in range(up, down+1):
                        res.append(matrix[i][right])
                    right -= 1
                if direct == 2:
                    for i in range(right, left-1, -1):
                        res.append(matrix[down][i])
                    down -= 1
                if direct == 3:
                    for i in range(down, up-1, -1):
                        res.append(matrix[i][left])
                    left += 1
                if up > down or left > right: 
                    return res
                direct = (direct+1) % 4

# Spiral Matrix II #

## 题目大意 ##

将一个正方形矩阵螺旋着填满递增的数字。

## 解题思路 ##

螺旋填满数字

## 代码 ##

class Solution(object):
        def generateMatrix(self, n):
            """
            :type n: int
            :rtype: List[List[int]]
            """
            if n == 0: 
                return []
            matrix = [[0 for i in range(n)] for j in range(n)]  # 创建矩阵
            up = 0
            down = len(matrix)-1
            left = 0
            right = len(matrix[0])-1
            direct = 0
            count = 0
            while True:
                if direct == 0:
                    for i in range(left, right+1):
                        count += 1; matrix[up][i] = count
                    up += 1
                if direct == 1:
                    for i in range(up, down+1):
                        count += 1; matrix[i][right] = count
                    right -= 1
                if direct == 2:
                    for i in range(right, left-1, -1):
                        count += 1; matrix[down][i] = count
                    down -= 1
                if direct == 3:
                    for i in range(down, up-1, -1):
                        count += 1; matrix[i][left] = count
                    left += 1
                if count == n*n: 
                    return matrix
                direct = (direct+1) % 4

## 总结 ##