[LeetCode] 402. Remove K Digits 去掉K位数字

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.

Java:

public String removeKdigits(String num, int k) {
    if(num.length()==k)
        return "0";
 
    StringBuilder sb = new StringBuilder(num);
    for(int j=0; j<k; j++){
        int i=0;
        while(i<sb.length()-1&&sb.charAt(i)<=sb.charAt(i+1)){
            i++;
        }
        sb.delete(i, i+1);
    }
 
    //remove leading 0's        
    while (sb.length() > 1 && sb.charAt(0)=='0')
        sb.delete(0,1);
 
    if(sb.length()==0){
        return "0";
    }
 
    return sb.toString();
}　　

Python:

class Solution(object):
    def removeKdigits(self, num, k):
        """
        :type num: str
        :type k: int
        :rtype: str
        """
        result = []
        for d in num:
            while k and result and result[-1] > d:
                result.pop()
                k -= 1
            result.append(d)
        return ''.join(result).lstrip('0')[:-k or None] or '0'

C++:

class Solution {
public:
    string removeKdigits(string num, int k) {
        string res = "";
        int n = num.size(), keep = n - k;
        for (char c : num) {
            while (k && res.size() && res.back() > c) {
                res.pop_back();
                --k;
            }
            res.push_back(c);
        }
        res.resize(keep);
        while (!res.empty() && res[0] == '0') res.erase(res.begin());
        return res.empty() ? "0" : res;
    }
};

　

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