LeetCode 402. Remove K Digits

402. Remove K Digits

Description
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.

Analysis
这道题的意思是求所给字符串删除k个字符后所得的最小整数。
我采用的方法是设置一个新的字符串s，同时将字符串遍历一遍。
当发现所给字符串后一个数字小于前面的数字时，即小于栈顶的数时，
我们将栈顶pop，同时继续比较，直到当前数字大于等于栈顶。
将当前数字push进s。
然而，这样做仍可能有问题，因为这个时候可能因为前面出现了当前数与前面的数字存在相等的情况，而导致错误。
所以还需要resize将s变成结果。

Code

class Solution {
public:
    string removeKdigits(string num, int k) {
        if(num.size() == 0 || num.size() <= k) return "0";
        if( k == 0) return num;
        int temp = num.size() - k;

        string s = "";
        for(int i = 0; isize();++i){
            while(k!= 0 && !s.empty() && num[i] < s.back()){
                s.pop_back();
                k--;
            }
            s.push_back(num[i]);
        }

        //s.resize(num.size() - k);
        s.resize(temp);
        while(!s.empty() && s[0] == '0') s.erase(s.begin()) ;

        if(s.size() == 0)  s+="0";
        return s;
    }     
};