Leetcode题解---滑动窗口

239. 滑动窗口最大值

动态规划 

class Solution:
    def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
        n = len(nums)
        if n * k == 0:
            return []
        if k == 1:
            return nums
        
        left = [0] * n
        left[0] = nums[0]
        right = [0] * n
        right[n - 1] = nums[n - 1]
        for i in range(1, n):
            # from left to right
            if i % k == 0:
                # block start
                left[i] = nums[i]
            else:
                left[i] = max(left[i - 1], nums[i])
            # from right to left
            j = n - i - 1
            if (j + 1) % k == 0:
                # block end
                right[j] = nums[j]
            else:
                right[j] = max(right[j + 1], nums[j])
        
        output = []
        for i in range(n - k + 1):
            output.append(max(left[i + k - 1], right[i]))
            
        return output

53. 最大子序和

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        dp=[0]*(len(nums))
        dp[0]=nums[0]
        for i in range(1,len(nums)):
            if dp[i-1]<=0:
                dp[i]=nums[i]
            else:
                dp[i]=dp[i-1]+nums[i]
        return max(dp)

 分治：

其实就是它的最大子序和要么在左半边，要么在右半边，要么是穿过中间，对于左右边的序列，情况也是一样，因此可以用递归处理。中间部分的则可以直接计算出来，时间复杂度应该是 O(nlogn)

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        n = len(nums)
        #递归终止条件
        if n == 1:
            return nums[0]
        else:
            #递归计算左半边最大子序和
            max_left = self.maxSubArray(nums[0:len(nums) // 2])
            #递归计算右半边最大子序和
            max_right = self.maxSubArray(nums[len(nums) // 2:len(nums)])
        
        #计算中间的最大子序和，从右到左计算左边的最大子序和，从左到右计算右边的最大子序和，再相加
        max_l = nums[len(nums) // 2 - 1]
        tmp = 0
        for i in range(len(nums) // 2 - 1, -1, -1):
            tmp += nums[i]
            max_l = max(tmp, max_l)
        max_r = nums[len(nums) // 2]
        tmp = 0
        for i in range(len(nums) // 2, len(nums)):
            tmp += nums[i]
            max_r = max(tmp, max_r)
        #返回三个中的最大值
        return max(max_right,max_left,max_l+max_r)