leetcode 之Spiral Matrix I 和 II 解题思路

题目如下：

Spiral Matrix

  My Submissions

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].

题意：就是按照螺旋形状打印出二维数组。

螺旋形状需要有四个方向，右，下，左，上，再依次进行。当在向右移动时，遇到边界或者后面一个已经打印过，则改变方向，向下；其它方向依次相同。因此需要有一个二维数组记录当前的位置是否打印过，还需要有一个变量记录已经打印了多少个，当打印的个数和二维数组总个数相同时，则停止打印。

代码如下：

public class Solution {
    public List spiralOrder(int[][] matrix) {
        List list = new ArrayList();
        int nRows = matrix.length;
        if(nRows == 0) return list;
        int nCols = matrix[0].length;
        int total_Len = nRows * nCols;//数组总长度
        int[][] used = new int[nRows][nCols];//用于判断matrix数组中的元素是否打印过
        
        int current_DIR = 0;//当前方向
        int i = 0; int j = 0;
        int len = 0;
        while(true){
        	if(len == total_Len){//如果当前打印长度和总长度相等，则停止打印
        		break;
        	}
        	switch(current_DIR){
        	case 0:{//0代表向右的方向
        		if(j == nCols || used[i][j] == 1){//改变方向的边界条件
        			j--;
        			i++;
        			current_DIR = 1;
        		}else{
        			list.add(matrix[i][j]);
        			used[i][j] = 1;
        			len++;
        			j++;
        		}
        		break;
        	}
        	case 1:{//1代表向下
        		if(i == nRows || used[i][j] == 1){
        			i--;
        			j--;
        			current_DIR = 2;
        		}else{
        			list.add(matrix[i][j]);
        			used[i][j] = 1;
        			len++;
        			i++;
        		}
        		break;
        	}
        	case 2:{//2代表向左
        		if(j == -1 || used[i][j] == 1){
        			j++;
        			i--;
        			current_DIR = 3;
        		}else{
        			list.add(matrix[i][j]);
        			used[i][j] = 1;
        			len++;
        			j--;
        		}
        		break;
        	}
        	case 3:{//3代表向上
        		if(i == -1 || used[i][j] == 1){
        			i++;
        			j++;
        			current_DIR = 0;
        		}else{
        			list.add(matrix[i][j]);
        			used[i][j] = 1;
        			len++;
        			i--;
        		}
        	}
        	}
        }
        return list;
    }
}

Spiral Matrix II

  Total Accepted: 10563  Total Submissions: 34583 My Submissions

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 ]
]

该题是根据给定的n按照螺旋形状构造一个n*n的矩阵，方法和打印类似，直接给出代码

public class Solution {
    public int[][] generateMatrix(int n) {
         if(n == 0) return new int[0][0];
        int[][] matrix = new int[n][n];
        int len = 0;
        int total_Len = n * n;
        int current_Dir = 0;
        int[][] used = new int[n][n];
        int i = 0, j = 0;
        while(true){
        	if(len == total_Len){
        		break;
        	}
        	switch(current_Dir){
        	case 0:{
        		if(j == n || used[i][j] == 1){
        			j--;
        			i++;
        			current_Dir = 1;
        		}else{
        			matrix[i][j] = ++len;
        			used[i][j] = 1;
        			j++;
        		}
        		break;
        	}
        	case 1:{
        		if(i == n || used[i][j] == 1){
        			i--;
        			j--;
        			current_Dir = 2;
        		}else{
        			matrix[i][j] = ++len;
        			used[i][j] = 1;
        			i++;
        		}
        		break;
        	}
        	case 2:{
        		if(j == -1 || used[i][j] == 1){
        			j++;
        			i--;
        			current_Dir = 3;
        		}else{
        			matrix[i][j] = ++len;
        			used[i][j] = 1;
        			j--;
        		}
        		break;
        	}
        	case 3:{
        		if(i == -1 || used[i][j] == 1){
        			i++;
        			j++;
        			current_Dir = 0;
        		}else{
        			matrix[i][j] = ++len;
        			used[i][j] = 1;
        			i--;
        		}
        		break;
        	}
        	}
        	
        }
        return matrix;
    }
}