题目：原题链接（中等）
标签：贪心算法、双指针、二分查找
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( N l o g N ) | O ( N ) | 544ms (29.16%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一：
class Solution:
def minOperations(self, nums: List[int], x: int) -> int:
# 寻找处理方案
last = 0
count_left = {0: 0}
for i, n in enumerate(nums):
last += n
count_left[last] = i + 1
# print(count_left)
choose = []
last = 0
for i, n in enumerate(nums[::-1]):
aim = x - last
if aim in count_left:
if count_left[aim] + i <= len(nums):
choose.append((count_left[aim], i))
last += n
# 寻找操作数最少的处理方案
if not choose:
return -1
choose.sort(key=lambda k: k[0] + k[1])
# print(choose)
# 处理字符串并返回结果
ans = choose[0]
nums[:] = nums[ans[0]:-ans[1]]
# print(nums)
return ans[0] + ans[1]
