Remove K Digits

problem:

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.

thinking:

采用贪心算法，要使得表示的数字最小，我们总使得高数位是小的数。从字符串起始位置开始，当后一个字符比前一个字符小，考虑去除前一个字符，保证前面的高数位是小的数。如果后一个字符都比前一个字符大，则考虑去除字符串末尾的数。我们需要去除字符串初始位置是’0’的字符。
算法时间为遍历一次字符串，复杂度为O(N)。

solution:

class Solution {  public:      string removeKdigits(string num, int k) {          for(int i=0;i<num.size()-1;i++)          {              if(k==0) break;              if(num[i]>num[i+1])               {                  num.erase(i,1);                  k--;                  i-=2;                  if(i<-1) i=-1;              }          }          if(k>0)          {              num.erase(num.size()-k,k);          }          int i=0;          while(true)          {              if(num[i]!='0') break;              i++;          }          num.erase(0,i);          if(num.size()==0) return "0";          return num;      }  };