题目：原题链接（困难）
标签：数组、贪心算法、动态规划
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( N ) | O ( N ) | 384ms (91.51%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
class Solution:
def maxProfit(self, prices: List[int]) -> int:
# 计算前缀最小值
prefix_min = []
last = (float("inf"), -1)
for i, price in enumerate(prices):
if price <= last[0]:
last = (price, i)
prefix_min.append(last)
# print(prefix_min)
# 寻找最大增幅段落
max_idx, max_val = [], 0
for i, price in enumerate(prices):
val = price - prefix_min[i][0]
if val > max_val:
max_idx, max_val = [(prefix_min[i][1], i)], val
elif val == max_val:
max_idx.append((prefix_min[i][1], i))
# print(max_idx, max_val)
# 如果有两段最大增幅且起点不同，那么他们一定不重叠
if len(set(x for x, y in max_idx)) > 1:
return 2 * max_val
max_i1, max_i2 = max_idx[0]
# 剩下的最大值要不存在于增幅段落中（反向），要不存在于增幅段落外（正向）
# 寻找存在于增幅段落中（反向）的最大值
prefix_max = []
last = float("-inf")
for i in range(max_i1, max_i2):
last = max(last, prices[i])
prefix_max.append(last)
max_v1 = 0
for i in range(max_i1, max_i2):
max_v1 = max(max_v1, prefix_max[i - max_i1] - prices[i])
# print("MAX1:", prefix_max, "->", max_v1)
# 寻找存在于增幅段落左侧（正向）的最大值
max_v2 = 0
for i in range(0, max_i1):
max_v2 = max(max_v2, prices[i] - prefix_min[i][0])
# print("MAX2:", max_v2)
# 寻找存在于增幅段落右侧（正向）的最大值
prefix_min = []
last = float("inf")
for i in range(max_i2, len(prices)):
last = min(last, prices[i])
prefix_min.append(last)
max_v3 = 0
for i in range(max_i2, len(prices)):
max_v3 = max(max_v3, prices[i] - prefix_min[i - max_i2])
# print("MAX3:", prefix_min, "->", max_v3)
return max_val + max(max_v1, max_v2, max_v3)
