给你一个 m 行 n 列的矩阵 matrix ,请按照 顺时针螺旋顺序 ,返回矩阵中的所有元素。
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]
提示:
m == matrix.lengthn == matrix[i].length1 <= m, n <= 10-100 <= matrix[i][j] <= 100
/**
* @param {number[][]} matrix
* @return {number[]}
*/
var spiralOrder = function(matrix) {
// let m = matrix.size()
// let n = matrix[0].size()
let up = left = 0
let down = right = matrix[0].length-1
let res = []
//let numCount = matrix.length*
while(true){
//遍历行
for(let i = left; i<= right; i++){
res.push(matrix[up][i])
}
++up
if(up > down){
break
}
//遍历右边的列
for(let i = up; i <= down; i++ ){
res.push(matrix[i][right])
}
--right
if(right < left){
break
}
//遍历下边的行
for(let i = right; i >= left; i--){
res.push(matrix[down][i])
}
if(right < left){
break
}
--down
//遍历左边的列
for(let i = down; i >= up; i--){
res.push(matrix[i][left])
}
++left
if(left > right){
break
}
}
return res
};