Leetode402. Remove K Digits-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_26621481/article/details/53040499

本文介绍了解决LeetCode 402题的方法，该题要求从一个非负整数中移除k位数字，使剩下的数字尽可能小。通过逐位检查并移除首位递减的数字来实现最小化。

LeetCode402. Remove K Digits

原题地址

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.

思路

从原数字的高位到低位检索，删除第一个下降的数字，然后又从头开始检索，依次删除K次，因为每删除一次数量级固定下降一位，所以如果是后一位比这一位大，删除这一位那么后一位更大的数就会在比较高的数量级上，但如果后一位更小的话，就会让小的数到高位上来。所以按照这个规律，最后在注意一下去掉开头的0，就可以得到结果。

代码实现

class Solution {
public:
    string remove(string & num, int pos) {
    return num.substr(0, pos) + num.substr(pos+1, num.size()-pos-1);
   }

    string removeKdigits(string num, int k) {

        if(num.size()<=k) return "0";

        for(int i=0; i<k; i++){
            for(int j=0; j<num.size(); j++){
                if(j==num.size()-1 || num[j]>num[j+1]){
                    num=remove(num,j);
                    break;
                }
            }
        }

        while(num[0]=='0') num=remove(num,0);
        if (num.size() == 0) return "0";//num="10",k=1的情况
        return num;
    }
};