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402. Remove K Digits

编程语言 mmisland 10℃ 0评论

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.

首先,利用贪婪算法的思想,将去掉k次最小转化为连续去掉k次1个数字,每次都最小。去掉1个数字最小也就是去掉从左向右检索的第一个产生降序的数字,即产生最小结果。比如1213,去掉2结果最小。整体思路是利用一个栈存储每一位数字,判定到达产生降序的数字时,就去掉该数字,算作去掉1个数字,此时再检查当前栈顶元素是不是新的产生降序的数字。直到k个数字删除结束。

string removeKdigits(string num, int k) {


        string res;


        int keep = num.size() – k;


        for (int i=0; i

            while (res.size()>0 && res.back()>num[i] && k>0) {


                res.pop_back();


                k–;


            }


            res.push_back(num[i]);


        }


        res.erase(keep, string::npos);


       


        // trim leading zeros


        int s = 0;


        while (s<(int)res.size()-1 && res[s]=='0')  s++;


        res.erase(0, s);


       


        return res==”” ? “0” : res;


    }

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