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。
示例一:
输入:matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出:[1,2,3,6,9,8,7,4,5]
实例二:
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.length
n == matrix[i].length
1 <= m, n <= 10
-100 <= matrix[i][j] <= 100
题解
此题和59一样,模拟就行了,没啥难的。注意边界的判断!
JavaScript
/**
* @param {number[][]} matrix
* @return {number[]}
*/
var spiralOrder = function(matrix) {
ans = []
let top = 0
let left = 0
let right = matrix[0].length-1
let bottom = matrix.length-1
let num = 1
let target = matrix[0].length*matrix.length
while(num<=target){
for(let i=left;i<right+1;i++){
if(num>target) continue
ans.push(matrix[top][i])
num++
}
top++
for(let i=top;i<bottom+1;i++){
if(num>target) continue
ans.push(matrix[i][right])
num++
}
right--
for(let i=right;i>left-1;i--){
if(num>target) continue
ans.push(matrix[bottom][i])
num++
}
bottom--
for(let i=bottom;i>top-1;i--){
if(num>target) continue
ans.push(matrix[i][left])
num++
}
left++
}
return ans
};
let m = [[1,2,3],[4,5,6],[7,8,9]]
let matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
console.log(spiralOrder(matrix))