打卡记录
相邻字符不同的最长路径（树状DP）
链接
若节点也存在父节点的情况下，传入父节点参数，若是遍历到父节点，直接循环里 continue。
class Solution:def longestPath(self, parent: List[int], s: str) -> int:n = len(parent)g = [[] for _ in range(n)]for i in range(1, n): g[parent[i]].append(i)ans = 0def dfs(x):nonlocal ansmax_len = 0for y in g[x]:length = dfs(y) + 1if s[x] != s[y]:ans = max(ans, max_len + length)max_len = max(max_len, length)return max_lendfs(0)return ans + 1
三个无重叠子数组的最大和（滑动窗口）
链接
分段处理，类似于股票问题的一种解法，把前面出现的最大值保存下来留给后面计算来使用，这样可以避免重叠出现。
class Solution:def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:ans = []sum1, maxSum1, maxSum1Idx = 0, 0, 0sum2, maxSum12, maxSum12Idx = 0, 0, ()sum3, maxTotal = 0, 0for i in range(k * 2, len(nums)):sum1 += nums[i - k * 2]sum2 += nums[i - k]sum3 += nums[i]if i >= k * 3 - 1:if sum1 > maxSum1:maxSum1 = sum1maxSum1Idx = i - k * 3 + 1if maxSum1 + sum2 > maxSum12:maxSum12 = maxSum1 + sum2maxSUm12Idx = (maxSum1Idx, i - k * 2 + 1)if maxSum12 + sum3 > maxTotal:maxTotal = maxSum12 + sum3ans = [*maxSUm12Idx, i - k + 1]sum1 -= nums[i - k * 3 + 1]sum2 -= nums[i - k * 2 + 1]sum3 -= nums[i - k + 1]return ans
三个无重叠子数组的最大和（动态规划）
链接
DP解法的难点在于如何回溯动态规划后的坐标，而且要保持字典序最小。
class Solution:def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:n = len(nums)s = list(accumulate(nums, initial=0))f = [[0] * 4 for _ in range(n + 2)]for i in range(n - k + 1, 0, -1):for j in range(1, 4):f[i][j] = max(f[i + 1][j], f[i + k][j - 1] + s[i + k - 1] - s[i - 1])ans = [0] * 3i, j, idx = 1, 3, 0while j > 0:if f[i + 1][j] > f[i + k][j - 1] + s[i + k - 1] - s[i - 1]: i += 1else:ans[idx] = i - 1idx += 1i += kj -= 1return ans
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