LeetCode-Remove K Digits

题目 Remove K Digits

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.

即给定一个非负数值字符串，要求删除k个字符的情况下数值最小。

测试例子

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.

代码

第一版本

/*************************************************************************> File Name: Solution.cpp> Author: Duke-wei> Mail: 13540639584@163.com > Created Time: 2017年10月13日 星期五 15时12分09秒************************************************************************/#include<iostream>
#include<string>
#include<stack>
#include<algorithm>
#include<vector>
using namespace std;
class Solution1{
public://首先简化问题，考虑一次去除一个字符的情况，然后去除k次string removeKdigits(string num, int k) {if(k==num.size()) return "0";for(int i=0;i<k;++i){if(num.size()<=1) return "0";num = removeone(num);//去掉头部0int j = 0;for(;j<num.size()&&num[j]=='0';++j);num = num.substr(j);}if(num.size()==0) return "0";return num;}//一次去除一个字符，通过找规律发现//去除一个字符，即开头非递减序列的最后一个//如1354278 即去除5,5是首个开始递减的字符//如12477589 去除7//如7511 去除7string removeone(string& s){int i = 0;for(;i<s.size()-1;++i){if(s[i]>s[i+1]) break;}string ret = s.substr(0,i);ret += s.substr(i+1);return ret;}
};

第二版本

每次删除一个肯定很慢，所有要遍历一次删除k个，发现这k个字符的规律，就可以通过栈来实现删除，从头入栈，发现比栈顶小，则弹出栈顶，即去除一个字符。

class Solution2{public:string removeKdigits(string num, int k) {if(k==num.size()) return "0";stack<char> sc;int i=0;while(k>0){if(i>=num.size()){sc.pop();--k;}else if(sc.empty()||sc.top()<=num[i]){sc.push(num[i++]);}else{sc.pop();--k;}}while(i<num.size()) sc.push(num[i++]);string ret;while(!sc.empty()){ret.push_back(sc.top());sc.pop();}reverse(ret.begin(),ret.end());i=0;while(i<ret.size()&&ret[i]=='0')++i;if(i==ret.size()) return "0";return ret.substr(i);}
};

第三版本

简化代码，看了大神的代码，简洁，直接用striing替换stack的功能。

class Solution3{public:string removeKdigits(string num, int k) {string ans = "";                                         // treat ans as a stack in below for loopfor (char c : num) {while (ans.length() && ans.back() > c && k) {ans.pop_back();                                  // make sure digits in ans are in ascending orderk--;                                             // remove one char}if (ans.length() || c != '0') { ans.push_back(c); }  // can't have leading '0'}while (ans.length() && k--) { ans.pop_back(); }          // make sure remove k digits in totalreturn ans.empty() ? "0" : ans;}
};

思考扩展

这题明确了该字符串是非负数值，因此如果数值为负则应该怎么处理？
如果要求去除k个后是最大怎么处理？