长度最小的子数组

• https://leetcode.cn/problems/minimum-size-subarray-sum/

思路

• 使用滑动窗口，left表示滑动窗口的起始点，right表示滑动窗口的终点

class Solution:def minSubArrayLen(self, target: int, nums: List[int]) -> int:left = 0     # left表示滑动窗口的起始点res = float('inf')   # float('inf')表示无穷大total = 0    # 记录滑动窗口中的和length = len(nums)for right in range(length):   # right表示滑动窗口的终点total += nums[right]while total >= target:l = right - left + 1res = min(l, res)total -= nums[left]left += 1if res == float('inf'):   # 说明在遍历过程中，初始的res始终没有被替换，即nums的总和小于targetreturn 0else:                # res被替换了，说明有满足条件的return res

螺旋矩阵

• https://leetcode.cn/problems/spiral-matrix-ii/

思路

• 每遍历一条边时，保持循环不变量原则，即每一条边都遵循左闭右开[)的原则

class Solution:def generateMatrix(self, n: int) -> List[List[int]]:arr = [ [None]*n for _ in range(n)]   # 先申请好一段二维数组startx, starty = 0, 0   # 每遍历一圈的起始位置offset = 1               # 右开count = 1               # 技术器loop = n // 2           # 总共螺旋的圈数mid = n // 2             # 若n为奇数，则最中间元素的坐标while loop > 0:for y in range(starty, n - offset):    # 上边arr[startx][y] = countcount+=1for x in range(startx, n - offset):   # 右边arr[x][n - offset] = countcount+=1for y in range(n - offset, starty, -1):   # 下边arr[n - offset][y] = countcount+=1for x in range(n - offset, startx, -1):   # 左边arr[x][starty] = countcount+=1startx += 1starty += 1offset += 1loop -= 1if n%2 == 1:   # 若n为奇数arr[mid][mid] = n*nreturn arr

总结