25期代码随想录算法训练营第二天 | 977.有序数组的平方 ，209.长度最小的子数组 ，59.螺旋矩阵II

977.有序数组的平方

链接

思路

双指针。

数组平方的最大值就在数组的两端，不是最左边就是最右边。所以我们可以用一左一右两个指针，来获取最大值，再放入到我们的结果集。

代码

class Solution:
    def sortedSquares(self, nums: List[int]) -> List[int]:
        l, r, i = 0, len(nums) - 1, len(nums) - 1
        res = [0 for _ in range(len(nums))]
        while l <= r:
            if nums[l] ** 2 < nums[r] ** 2:
                res[i] = nums[r] ** 2
                r -= 1
            else:
                res[i] = nums[l] ** 2
                l += 1
            i -= 1
        return res
1
2
3
4
5
6
7
8
9
10
11
12
13

209.长度最小的子数组

链接

滑动窗口

代码

class Solution:
    def sortedSquares(self, nums: List[int]) -> List[int]:
        l, r, i = 0, len(nums) - 1, len(nums) - 1
        res = [0 for _ in range(len(nums))]
        while l <= r:
            if nums[l] ** 2 < nums[r] ** 2:
                res[i] = nums[r] ** 2
                r -= 1
            else:
                res[i] = nums[l] ** 2
                l += 1
            i -= 1
        return res
1
2
3
4
5
6
7
8
9
10
11
12
13

59.螺旋矩阵II

链接

代码

class Solution:
    def generateMatrix(self, n: int) -> List[List[int]]:
        res = [[0] * n for _ in range(n)]
        startx, starty = 0, 0
        loop, mid = n // 2, n // 2
        count = 1
        
        for offset in range(1, loop + 1):
            #upper left to right
            for i in range(starty, n - offset):
                res[startx][i] = count
                count += 1
            
            #up to bottom
            for i in range(startx, n - offset):
                res[i][n - offset] = count
                count += 1

            # from bottom right to left
            for i in range(n - offset, starty, -1):
                res[n - offset][i] = count
                count += 1
            
            # from bottom to up
            for i in range(n - offset, startx, -1):
                res[i][starty] = count
                count += 1
            
            startx += 1
            starty += 1
        
        if n % 2 != 0:
            res[mid][mid] = count

        return res

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36