LeetCode689 三个无重叠子数组的最大和

LeetCode689 三个无重叠子数组的最大和

$dp[i][j]$表示前$j$个元素，选$i$个子区间的最优值

对于当前$j$，转移方程考虑是否以第$j$个元素为子区间的最后一个元素

$dp[i][j] = \max(dp[i][j - 1], dp[i - 1][j - k] + pre[j] - pre[j - k])$

class Solution:
    def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:

        n, pre = len(nums), [0]
        for i in range(n): pre.append(pre[-1] + nums[i])
        dp = [[0 for i in range(n + 1)] for j in range(3 + 1)]

        for i in range(1, 3 + 1):
            for j in range(k, n + 1):
                dp[i][j] = max(dp[i][j - 1], dp[i - 1][j - k] + pre[j] - pre[j - k])
        
        # for i in range(3 + 1): print(dp[i])
        ans, j, i = [], 3, n
        while i > 0:
            if j == 0: break
            if dp[j][i] > dp[j][i - 1]:
                ans.append(i - k)
                j = j - 1
                i = i - k
            else:
                i = i - 1

        ans.reverse()
        return ans