JavaScript / TypeScript for LeetCode
当前进程：
开始时间：2020.6.27结束时间：undefined
GitHub仓库：https://github.com/Cundefined/JavaScript-or-TypeScript-for-LeetCode
参考视频：https://www.bilibili.com/video/BV1wA411b7qZ
1、题目要求
( LeetCode-第54题 ) 螺旋矩阵 给定一个包含 m x n 个元素的矩阵（m 行, n 列），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
:
[
[ 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
:
[
[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]
来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/spiral-matrix 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
2、解题思路
判断输入是否为空，为空就返回空数组
用left、right、top、bottom定义4个边界
初始化遍历方向，依次是向右、下、左、上，顺时针螺旋
初始化结果数组
while循环，只要左边界小于等于右边界，并且上边界小于等于下边界，就一直螺旋遍历：
5.1、依次按照向右、下、左、上的方向遍历二维数组，并且将沿途的元素保存在结果数组
返回结果数组
2.1、JavaScript Solution
var spiralOrder = function (matrix
) {
if (matrix
.length
=== 0) {
return [];
}
let top
= 0;
let right
= matrix
[0].length
- 1;
let bottom
= matrix
.length
- 1;
let left
= 0;
let direction
= "right";
let result
= [];
while (left
<= right
&& top
<= bottom
) {
if (direction
=== "right") {
for (let i
= left
; i
<= right
; i
++) {
result
.push(matrix
[top
][i
]);
}
top
++;
direction
= "bottom";
} else if (direction
=== "bottom") {
for (let i
= top
; i
<= bottom
; i
++) {
result
.push(matrix
[i
][right
]);
}
right
--;
direction
= "left";
} else if (direction
=== "left") {
for (let i
= right
; i
>= left
; i
--) {
result
.push(matrix
[bottom
][i
]);
}
bottom
--;
direction
= "top";
} else if (direction
=== "top") {
for (let i
= bottom
; i
>= top
; i
--) {
result
.push(matrix
[i
][left
]);
}
left
++;
direction
= "right";
}
}
return result
;
};
2.2、TypeScript Solution
function spiralOrder(matrix
: number
[][]): number
[] {
if (matrix
.length
=== 0) {
return [];
}
let top
: number
= 0;
let right
: number
= matrix
[0].length
- 1;
let bottom
: number
= matrix
.length
- 1;
let left
: number
= 0;
let direction
: string
= "right";
let result
: number
[] = [];
while (left
<= right
&& top
<= bottom
) {
if (direction
=== "right") {
for (let i
: number
= left
; i
<= right
; i
++) {
result
.push(matrix
[top
][i
]);
}
top
++;
direction
= "bottom";
} else if (direction
=== "bottom") {
for (let i
: number
= top
; i
<= bottom
; i
++) {
result
.push(matrix
[i
][right
]);
}
right
--;
direction
= "left";
} else if (direction
=== "left") {
for (let i
: number
= right
; i
>= left
; i
--) {
result
.push(matrix
[bottom
][i
]);
}
bottom
--;
direction
= "top";
} else if (direction
=== "top") {
for (let i
: number
= bottom
; i
>= top
; i
--) {
result
.push(matrix
[i
][left
]);
}
left
++;
direction
= "right";
}
}
return result
;
}
