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《明日歌》 ——钱福（明）
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世人若被明日累，春去秋来老将至。
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百年明日能几何？请君听我明日歌。
描述
Given string num representing a non-negative integer num, and an integer k, return the smallest possible integer after removing k digits from num.
Example 1:
Input: num = "1432219", k = 3
Output: "1219"
Explanation: Remove the three digits 4, 3, and 2 to form the new number 1219 which is the smallest.
Example 2:
Input: num = "10200", k = 1
Output: "200"
Explanation: Remove the leading 1 and the number is 200. Note that the output must not contain leading zeroes.
Example 3:
Input: num = "10", k = 2
Output: "0"
Explanation: Remove all the digits from the number and it is left with nothing which is 0.
Note:
1 <= k <= num.length <= 10^5
num consists of only digits.
num does not have any leading zeros except for the zero itself.
解析
根据题意，给定表示非负整数 num 的字符串 num 和整数 k，从 num 中删除 k 位后返回可能的最小整数。题意很简单，用到的思路也是单调栈解法，因为删除之后的数字想要保证最小整数，所以维护一个单调递增的栈 stack ，尽量将前面较大的数字删除。如果只删除了小于 k 的数字，说明 num 此时是一个递增的字符串，我们只需要将最后的几个数字删除即可。需要注意的是前置 0 要去掉，而且如果数字都被删除，结果应该是 "0" 。
解答
class Solution(object):
def removeKdigits(self, num, k):
"""
:type num: str
:type k: int
:rtype: str
"""
stack = []
for c in num:
while stack and k>0 and stack[-1]>c:
stack.pop()
k-=1
stack.append(c)
if k>0:
stack = stack[:-k]
result = "".join(stack).lstrip("0")
if not result:
return "0"
return result
运行结果
Runtime: 28 ms, faster than 84.06% of Python online submissions for Remove K Digits.
Memory Usage: 14.1 MB, less than 14.74% of Python online submissions for Remove K Digits.
原题链接：leetcode.com/problems/re…
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