Given a matrix of*m*x*n*elements (*m*rows,*n*columns), return all elements of the matrix in spiral order. **Example 1:** ~~~ Input: [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] Output: [1,2,3,6,9,8,7,4,5] ~~~ **Example 2:** ~~~ Input: [ [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] ~~~ 給定一個包含?m x n?個元素的矩陣（m 行, n 列），請按照順時針螺旋順序，返回矩陣中的所有元素。 示例?1: 輸入: ``` [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ] ] ``` 輸出: `[1,2,3,6,9,8,7,4,5]` 示例?2: 輸入: ``` [ [1, 2, 3, 4], [5, 6, 7, 8], [9,10,11,12] ] ``` 輸出: `[1,2,3,4,8,12,11,10,9,5,6,7]`  ``` /** * @param {number[][]} matrix * @return {number[]} */ var spiralOrder = function(matrix) { var res=[] var i=0 var j=0 var n=matrix.length-1 if(n<0) return [] var m=matrix[0].length-1 var turn=m==0?'d':'r' var boundl=0 var boundr=m var boundu=0 var boundd=n for(var a=0;a<(m+1)*(n+1);a++){ res.push(matrix[i][j]) if(turn=='r'){ j++ if(j==boundr){ boundu++ turn='d' } }else if(turn=='d'){ i++ if(i==boundd){ boundr-- turn='l' } }else if(turn=='l'){ j-- if(j==boundl){ boundd-- turn='u' } }else if(turn=='u'){ i-- if(i==boundu){ boundl++ turn='r' } } } return res }; ```