代码随想录算法训练营Day2|209.长度最小的子数组 59.螺旋矩阵

学习资料：https://programmercarl.com/数组总结篇.html#数组的经典题目

移动窗格，首尾指针根据条件变化
模拟行为，循环不变量（左闭右闭或左闭右开）整个过程保持一致

学习记录：
209.长度最小的子数组（用while使得尾指针遍历全部；用while实现，当[首:尾]之和>目标值，才移动首指针；为了求最小长度，先设置最小长度为正无穷float('inf')）

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class Solution(object):
    def minSubArrayLen(self, target, nums):
        """
        :type target: int
        :type nums: List[int]
        :rtype: int
        """
        begin=0
        end=0
        sum=0
        min_len=float('inf')   #设置为无穷大，方便找到最小值
        while end < len(nums):
            sum += nums[end]
            while sum >= target:
                sum -= nums[begin]
                sub_l=end-begin+1
                min_len = min(sub_l, min_len)
                begin += 1
            end += 1
        if min_len != float('inf'):
            return min_len
        else:
            return 0

59.螺旋矩阵(边界值很多，有初始横纵指针、左闭右开需要从n剪掉的offset、从1开始计数的count；当奇数阶矩阵，中间多出来的1项单独设置；螺旋转的圈数为n//2)

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class Solution(object):
    def generateMatrix(self, n):
        """
        :type n: int
        :rtype: List[List[int]]
        """
        start_x = 0
        start_y = 0
        nums = [[0]*n for _ in range(n)]
        cycle, mid = n//2, n//2
        offset=1
        count=1
        for c in range(cycle):
            for j in range(start_y, n-offset):
                nums[start_x][j]=count
                count += 1
            for i in range(start_x, n-offset):
                nums[i][n-offset]=count 
                count += 1
            for j in range(n-offset, start_y, -1):
                nums[n-offset][j]=count
                count += 1
            for i in range(n-offset, start_x, -1):
                nums[i][start_y] = count
                count += 1
            start_x += 1
            start_y += 1
            offset += 1
        if n%2 != 0:
            nums[mid][mid]=count
        return nums