[Leetcode] Spiral Matrix 螺旋矩阵

时间 2019-11-08

原文原文链接

Spiral Matrix I

Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order.python

For example, Given the following matrix:app
[
 [ 1, 2, 3 ],
 [ 4, 5, 6 ],
 [ 7, 8, 9 ]
]
You should return [1,2,3,6,9,8,7,4,5].code

顺序添加法

复杂度

时间 O(NM) 空间 O(1)element

思路

首先考虑最简单的状况，如图咱们先找最外面这圈X，这种状况下咱们是第一行找前4个，最后一列找前4个，最后一行找后4个，第一列找后4个，这里咱们能够发现，第一行和最后一行，第一列和最后一列都是有对应关系的。即对i行，其对应行是m - i - 1，对于第j列，其对应的最后一列是n - j - 1。it

XXXXX
XOOOX
XOOOX
XOOOX
XXXXX

而后进入到里面那一圈，一样的顺序没什么问题，然而关键在于下图这么两种状况，一圈已经没有四条边了，因此咱们要单独处理，只加那惟一的一行或一列。另外，根据下图咱们能够发现，圈数是宽和高中较小的那个，加1再除以2。io

OOOOO  OOO
OXXXO  OXO
OOOOO  OXO
       OXO
       OOO

代码

public class Solution {
    public List<Integer> spiralOrder(int[][] matrix) {
        List<Integer> res = new LinkedList<Integer>();
        if(matrix.length == 0) return res;
        int m = matrix.length, n = matrix[0].length;
        // 计算圈数
        int lvl = (Math.min(m, n) + 1) / 2;
        for(int i = 0; i < lvl; i++){
            // 计算相对应的该圈最后一行
            int lastRow = m - i - 1;
            // 计算相对应的该圈最后一列
            int lastCol = n - i - 1;
            // 若是该圈第一行就是最后一行，说明只剩下一行
            if(i == lastRow){
                for(int j = i; j <= lastCol; j++){
                    res.add(matrix[i][j]);
                }
            // 若是该圈第一列就是最后一列，说明只剩下一列
            } else if(i == lastCol){
                for(int j = i; j <= lastRow; j++){
                    res.add(matrix[j][i]);
                }
            } else {
                // 第一行
                for(int j = i; j < lastCol; j++){
                    res.add(matrix[i][j]);
                }
                // 最后一列
                for(int j = i; j < lastRow; j++){
                    res.add(matrix[j][lastCol]);
                }
                // 最后一行
                for(int j = lastCol; j > i; j--){
                    res.add(matrix[lastRow][j]);
                }
                // 第一列
                for(int j = lastRow; j > i; j--){
                    res.add(matrix[j][i]);
                }
            }
        }
        return res;
    }
}

2018/2ast

class Solution:
    def spiralOrder(self, matrix):
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        if (matrix is None or len(matrix) == 0):
            return []
        rows = len(matrix)
        cols = len(matrix[0])
        level = (min(rows, cols) + 1) // 2
        result = []
        for currLevel in range(0, level):
            firstRow = currLevel
            firstCol = currLevel
            lastRow = rows - currLevel - 1
            lastCol = cols - currLevel - 1
            if firstRow == lastRow:
                for col in range(firstCol, lastCol + 1):
                    result.append(matrix[firstRow][col])
                return result
            if firstCol == lastCol:
                for row in range(firstRow, lastRow + 1):
                    result.append(matrix[row][firstRow])
                return result
            for col in range(firstCol, lastCol):
                result.append(matrix[firstRow][col])
            for row in range(firstRow, lastRow):
                result.append(matrix[row][lastCol])
            for col in range(lastCol, firstCol, -1):
                result.append(matrix[lastRow][col])
            for row in range(lastRow, firstRow, -1):
                result.append(matrix[row][firstCol])
        return result

Spiral Matrix II

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

For example, Given n = 3,变量

You should return the following matrix:List
[
 [ 1, 2, 3 ],
 [ 8, 9, 4 ],
 [ 7, 6, 5 ]
]

顺序添加法

复杂度

时间 O(NM) 空间 O(1)

思路

本题就是按照螺旋的顺序把数字依次塞进去，咱们能够维护上下左右边界的四个变量，一圈一圈往里面添加。最后要注意的是，若是n是奇数，要把中心那个点算上。

代码

public class Solution {
    public int[][] generateMatrix(int n) {
        int[][] res = new int[n][n];
        int left = 0, right = n - 1, bottom = n - 1, top = 0, num = 1;
        while(left < right && top < bottom){
            // 添加该圈第一行
            for(int i = left; i < right; i++){
                res[top][i] = num++;
            }
            // 添加最后一列
            for(int i = top; i < bottom; i++){
                res[i][right] = num++;
            }
            // 添加最后一行
            for(int i = right; i > left; i--){
                res[bottom][i] = num++;
            }
            // 添加第一列
            for(int i = bottom; i > top; i--){
                res[i][left] = num++;
            }
            top++;
            bottom--;
            left++;
            right--;
        }
        // 若是是奇数，加上中间那个点
        if(n % 2 == 1){
            res[n / 2][n / 2] = num;
        }
        return res;
    }
}

2018/2

class Solution:
    def generateMatrix(self, n):
        """
        :type n: int
        :rtype: List[List[int]]
        """
        if n == 0:
            return []
        matrix = [[0 for i in range(0, n)] for j in range(0, n)]
        left, right, top, bottom, num = 0, n - 1, 0, n - 1, 1
        while left < right and top < bottom:
            for col in range(left, right):
                matrix[top][col] = num
                num += 1
            for row in range(top, bottom):
                matrix[row][right] = num
                num += 1
            for col in range(right, left, -1):
                matrix[bottom][col] = num
                num += 1
            for row in range(bottom, top, -1):
                matrix[row][left] = num
                num += 1
            left += 1
            right -= 1
            top += 1
            bottom -= 1
        if n % 2 != 0:
            matrix[left][top] = num
        return matrix

后续 Follow Up

若是在matrix ii中，给出的是m和n来表明行数和列数，该如何解决？

i和ii的本质区别就是一个是任意长方形，一个是正方形，因此ii中不须要判断最后一行或者最后一列。既然没有了正方形的前提，那生成矩阵的方法就和i是同样的了。

class Solution:
    def generateMatrix(self, m, n):
        # m rows, n cols
        if m == 0 or n == 0:
            return []
        matrix = [[0 for i in range(0, n)] for j in range(0, m)]
        num = 1
        level = (min(m, n) + 1) // 2
        for currLevel in range(0, level):
            lastRow = m - currLevel - 1
            lastCol = n - currLevel - 1
            firstRow = currLevel
            firstCol = currLevel
            if firstRow == lastRow:
                for col in range(firstCol, lastCol + 1):
                    matrix[firstRow][col] = num
                    num += 1
                return matrix
            if firstCol == lastCol:
                for row in range(firstRow, lastRow + 1):
                    matrix[row][firstRow] = num
                    num += 1
                return matrix
            for col in range(firstCol, lastCol):
                matrix[firstRow][col] = num
                num += 1
            for row in range(firstRow, lastRow):
                matrix[row][lastCol] = num
                num += 1
            for col in range(lastCol, firstCol, -1):
                matrix[lastRow][col] = num
                num += 1
            for row in range(lastRow, firstRow, -1):
                matrix[row][firstCol] = num
                num += 1
        return matrix