LeetCode #402 - Remove K Digits_要求删除其中的k位数字,在不改变数字原来的相对顺序的前提下,请告诉我们,剩下的数-优快云博客

本文链接：https://blog.youkuaiyun.com/LawFile/article/details/81231030

本文介绍了一个算法问题：如何从一个非负整数中移除k位数，使得剩下的数字尽可能小。通过实例说明了算法的实现过程，并提供了一段C++代码示例，展示了如何通过维护一个单调递增的数列来解决这个问题。

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题目描述：

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.

求把一个数中的k位数移除，可能得到的最小的数。可以移除k个数，那么在不改变每位数的相对顺序的前提下，可以尽量维持一个单增序列。只要还能继续删除数，就要尽量把较小的数放在高位，也就是删除它之前比它大的数。

class Solution {
public:
    string removeKdigits(string num, int k) {
        int n=num.size();
        if(n<=k) return "0";
        int count=k;
        vector<char> result;
        for(int i=0;i<n;i++)
        {
            while(result.size()>0&&count>0&&num[i]<result.back())
            {
                result.erase(result.end()-1);
                count--;
            }
            result.push_back(num[i]);
        }
        
        vector<char> x(result.begin(),result.begin()+n-k);
        while(x.size()>0&&x[0]=='0') x.erase(x.begin());
        if(x.size()==0) return "0";
        string s;
        for(int i=0;i<x.size();i++) s+=x[i];
        return s;
    }
};