顺时针打印矩阵(Print matrix clockwise)

输入一个矩阵,按照从外向里以顺时针的顺序依次打印出每一个数字。

示例 1:

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

输入:matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]输出:[1,2,3,4,8,12,11,10,9,5,6,7]

class Solution {
    public int[] spiralOrder(int[][] matrix) {
        if(matrix.length==0) return new int[0];
       

        int l=0,r=matrix[0].length-1,t=0,b=matrix.length-1,x=0;
         int[] res=new int[(r+1)*(b+1)];

        while(true){
        for(int i=l;i<=r;i++)
            res[x++]=matrix[t][i];
            if(++t>b)
            break;

        for(int i=t;i<=b;i++)
            res[x++]=matrix[i][r];
            if(l>--r)
            break;
        for(int i=r;i>=l;i--)
            res[x++]=matrix[b][i];
            if(t>--b)
            break;
         for(int i=b;i>=t;i--)
            res[x++]=matrix[i][l];
            if(++l>r)
            break;
        }
       
        return res;
    }
}
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Enter a matrix and print out each number in clockwise order from the outside to the inside.

Example 1:

Input: matrix = [[1,2,3], [4,5,6], [7,8,9]] output: [1,2,3,6,9,8,7,4,5] example 2:

Input: matrix = [[1,2,3,4], [5,6,7,8], [9,10,11,12]] output: [1,2,3,4,8,12,11,10,9,5,6,7]

class Solution {
    public int[] spiralOrder(int[][] matrix) {
        if(matrix.length==0) return new int[0];
       

        int l=0,r=matrix[0].length-1,t=0,b=matrix.length-1,x=0;
         int[] res=new int[(r+1)*(b+1)];

        while(true){
        for(int i=l;i<=r;i++)
            res[x++]=matrix[t][i];
            if(++t>b)
            break;

        for(int i=t;i<=b;i++)
            res[x++]=matrix[i][r];
            if(l>--r)
            break;
        for(int i=r;i>=l;i--)
            res[x++]=matrix[b][i];
            if(t>--b)
            break;
         for(int i=b;i>=t;i--)
            res[x++]=matrix[i][l];
            if(++l>r)
            break;
        }
       
        return res;
    }
}