题目
Given an integer array nums and a positive integer k, return the most competitive subsequence of nums of size k.
An array’s subsequence is a resulting sequence obtained by erasing some (possibly zero) elements from the array.
We define that a subsequence a is more competitive than a subsequence b (of the same length) if in the first position where a and b differ, subsequence a has a number less than the corresponding number in b. For example, [1,3,4] is more competitive than [1,3,5] because the first position they differ is at the final number, and 4 is less than 5.
解题思路
贪心，维护一个单调栈使得结果中的数是最小的，当剩余数的个数刚好能凑够k个的时候，就不能再pop了
代码
class Solution:
def mostCompetitive(self, nums: List[int], k: int) -> List[int]:
stack = []
for i, n in enumerate(nums):
while stack and stack[-1] > n and i + k - len(stack) < len(nums):
stack.pop()
if len(stack) < k:
stack.append(n)
return stack
