删除字符串中k个数字使获得的数字最小 Remove K Digits

问题：

Given a non-negative integer num represented as a string, remove k digits from the number so that the new number is the smallest possible.

Note:

• The length of num is less than 10002 and will be ≥ k.
• The given num does not contain any leading zero.

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.

解决：

① 

有一个N为长的数字，用字符串来代替了，如今要求你将它删除K位，使得其获得的结果最小。对于每次删除的状况有两种

(1)假如第一个数不为0，第二个数为0，那么咱们删除第一个数，就至关于数量级减小2，这样比删除获得的数其余任何一个方法都小

(2)另外一种状况，咱们找到第一次遍历的局部最大值，即遍历num第一个知足num.charAt(i)>num.charAt(i+1)的值，删除这个点，获得的值最小。这里就是贪心算法，每次删除一个局部最大。

class Solution {//6ms     public String removeKdigits(String num, int k) {         StringBuilder sb = new StringBuilder();         int len = num.length();         char[] stack = new char[len];         int count = 0;         for (int i = 0;i < len;i ++){             while(count != 0 && k > 0 && num.charAt(i) < stack[count - 1]){//根据贪心算法删除比后一个大的值                 count --;                 k --;             }             stack[count ++] = num.charAt(i);         }         int start = 0;         while(start < count && stack[start] == '0') start ++;         return start >= count - k ? "0" : new String(stack,start,count - start - k);     } }