leetcode-54-Spiral-Matrix

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描述

Given a positive integer n, generate a square matrix filled with elements from 1 to n2 in spiral order.

Example:

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Input: 3
Output:
[
 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
]

分析

模拟题，注意边界情况即可。

解决方案1(Python)

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class Solution(object):
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        result = []
        while matrix and matrix[0]:
            if matrix and matrix[0]:
                result += matrix.pop(0)
            if matrix and matrix[0]:
                for row in matrix:
                    result.append(row.pop())
            if matrix and matrix[0]:
                result += matrix.pop()[::-1]
            if matrix and matrix[0]:
                for row in matrix[::-1]:
                    result.append(row.pop(0))
        return result

解决方案2（Golang）

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func spiralOrder(matrix [][]int) []int {
    if len(matrix) == 0 || len(matrix[0]) == 0 {
        return []int{}
    }
    
    rows, columns := len(matrix), len(matrix[0])
    visited := make([][]bool, rows)
    for i := 0; i < rows; i++ {
        visited[i] = make([]bool, columns)
    }
    
    var (
        total = rows * columns
        result = make([]int, total)
        row, column = 0, 0
        directions = [][]int{[]int{0, 1}, []int{1, 0}, []int{0, -1}, []int{-1, 0}}
        directionIndex = 0
    )
    
    for i := 0; i < total; i++ {
        result[i] = matrix[row][column]
        visited[row][column] = true
        nextRow, nextColumn := row + directions[directionIndex][0], column + directions[directionIndex][1]
        if nextRow < 0 || nextRow >= rows || nextColumn < 0 || nextColumn >= columns || visited[nextRow][nextColumn] {
            directionIndex = (directionIndex+1) % 4
        }
        row += directions[directionIndex][0]
        column += directions[directionIndex][1]
    }
    return result
}

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