LeetCode.54(59) Spiral Matrix && II

题目54：

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

For example,
Given the following matrix:

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

You should return [1,2,3,6,9,8,7,4,5].

分析：

class Solution {    public List<Integer> spiralOrder(int[][] matrix) {        //给定矩阵，螺旋输出其所有元素        //思路：顺序为：右->下->左->上，每次走完一个方向，及时更新m或者n，就判断是否满足往下一个方向的行走        List<Integer> list=new ArrayList<>();        if(matrix.length==0){            return list;        }                int rowBegin=0,rowEnd=matrix.length-1;        int colBegin=0,colEnd=matrix[0].length-1;                //保证begin小于等于end        while(rowBegin<=rowEnd&&colBegin<=colEnd){            //右            for(int j=colBegin;j<=colEnd;j++){                list.add(matrix[rowBegin][j]);            }            //更新row            rowBegin++;                        //下            for(int i=rowBegin;i<=rowEnd;i++){                list.add(matrix[i][colEnd]);            }            //更新col            colEnd--;                        //判断是否有空间往回走            if(rowBegin<=rowEnd){                //左                for(int j=colEnd;j>=colBegin;j--){                    list.add(matrix[rowEnd][j]);                }            }            //更新            rowEnd--;                        //判断是否有空间往上走             if(colBegin<=colEnd){                 for(int i=rowEnd;i>=rowBegin;i--){                     list.add(matrix[i][colBegin]);                 }             }            //更新            colBegin++;         }        return list;             }}

题目(59):

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

For example,
Given n = 3,

You should return the following matrix:

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

分析：

class Solution {    public int[][] generateMatrix(int n) {        //类似sprial matrix给定矩阵，返回数据。该题给定数据，返回矩阵。                int [][] matrix=new int[n][n];        int rowBegin=0,rowEnd=n-1;        int colBegin=0,colEnd=n-1;                int count=1;                while(rowBegin<=rowEnd&&colBegin<=colEnd){            //右            for(int i=colBegin;i<=colEnd;i++){                matrix[rowBegin][i]=count;                count++;            }            rowBegin++;                        //下            for(int i=rowBegin;i<=rowEnd;i++){                matrix[i][colEnd]=count;                count++;            }            colEnd--;                //左            if(rowBegin<=rowEnd){                for(int i=colEnd;i>=colBegin;i--){                    matrix[rowEnd][i]=count;                    count++;                }            }            rowEnd--;                        //上            if(colBegin<=colEnd){                for(int i=rowEnd;i>=rowBegin;i--){                    matrix[i][colBegin]=count;                    count++;                }            }            colBegin++;        }        return matrix;            }}