有序数组的平法

题目描述：

调库解法：

class Solution(object):
    def sortedSquares(self, nums):
        square = sorted(map(lambda x: x*x, nums))
        return square

双指针解法：

class Solution(object):
    def sortedSquares(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        i = 0
        j = len(nums) - 1
        k = len(nums) - 1
        result = [0] * (k + 1)
        while i <= j:
            if nums[i] * nums[i] < nums[j] * nums[j]:
                result[k] = nums[j] * nums[j]
                k -= 1
                j -= 1
            if nums[i] * nums[i] >= nums[j] * nums[j]:
                result[k] = nums[i] * nums[i]
                k -= 1
                i += 1        
        return result

长度最小的子数组

题目描述：

双指针解法：

class Solution(object):
    def minSubArrayLen(self, target, nums):
        """
        :type target: int
        :type nums: List[int]
        :rtype: int
        """
        sum = 0
        i = 0
        j = 0
        result = float("inf")
        subLength = 0
        for j in range(len(nums)):
            sum += nums[j]
            while sum >= target:
                subLength = j - i + 1
                result = min(result, subLength)
                sum -= nums[i]
                i += 1
        return 0 if result == float("inf") else result

螺旋矩阵Ⅱ

题目描述：

解题代码：

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