class Solution:
def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
res = []
while matrix:
res += matrix.pop(0)
matrix = list(map(list, zip(*matrix)))[::-1]
return res
Runtime: 20 ms, faster than 100.00% of Python3 online submissions for Spiral Matrix.
Memory Usage: 12.8 MB, less than 100.00% of Python3 online submissions for Spiral Matrix.
方法还是绕着外面一点点算进去。。
var spiralOrder = function(matrix) {
if (matrix.length === 0) {
return []
}
const ans = []
let u = 0,
d = matrix.length - 1,
l = 0,
r = matrix[0].length - 1
while (1) {
for (let i = l; i <= r; ++i) {
ans.push(matrix[u][i])
}
if (++u > d) {
break
}
for (let i = u; i <= d; ++i) {
ans.push(matrix[i][r])
}
if (--r < l) {
break
}
for (let i = r; i >= l; --i) {
ans.push(matrix[d][i])
}
if (--d < u) {
break
}
for (let i = d; i >= u; --i) {
ans.push(matrix[i][l])
}
if (++l > r) {
break
}
}
return ans
};
Runtime: 48 ms, faster than 88.01% of JavaScript online submissions for Spiral Matrix.
Memory Usage: 33.8 MB, less than 36.36% of JavaScript online submissions for Spiral Matrix.
var spiralOrder = function(matrix) {
if (matrix.length === 0) {
return []
}
const list = []
let m = matrix.length
n = matrix[0].length
i = 0
count = ~~((Math.min(m, n) + 1) / 2)
while (i < count) {
for (let j = i; j < n - i; j++) {
list.push(matrix[i][j]);
}
for (let j = i + 1; j < m - i; j++) {
list.push(matrix[j][(n - 1) - i]);
}
for (let j = (n - 1) - (i + 1); j >= i && (m - 1 - i != i); j--) {
list.push(matrix[(m - 1) - i][j]);
}
for (let j = (m - 1) - (i + 1); j >= i + 1 && (n - 1 - i) != i; j--) {
list.push(matrix[j][i]);
}
i++;
}
return list
};
Runtime: 52 ms, faster than 68.49% of JavaScript online submissions for Spiral Matrix.
Memory Usage: 33.8 MB, less than 63.64% of JavaScript online submissions for Spiral Matrix.