题目:原题链接(困难)
标签:动态规划、二分查找
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | 116ms (6.07%) | ||
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一:
class Solution:
def splitArray(self, nums: List[int], m: int) -> int:
if m == 1:
return sum(nums)
size = len(nums)
# dp[i][j] 前j个字符分为i个子数组的最小最大值
dp = [[0] * len(nums) for _ in range(m)]
# dp[0]可以当做前缀和使用
dp[0][0] = nums[0]
for j in range(1, size):
dp[0][j] = dp[0][j - 1] + nums[j]
for i in range(1, m):
j1 = i - 1
for j2 in range(i, size):
dp[i][j2] = max(dp[i - 1][j1], dp[0][j2] - dp[0][j1])
while dp[i - 1][j1 + 1] <= dp[0][j2] - dp[0][j1]:
j1 += 1
dp[i][j2] = min(dp[i][j2], max(dp[i - 1][j1], dp[0][j2] - dp[0][j1]))
# for row in dp:
# print(row)
return dp[-1][-1]