[LeetCode]Delete Digits

题目

Given string A representative a positive integer which has N digits, remove any k digits of the number, the remaining digits are arranged according to the original order to become a new positive integer.
Find the smallest integer after remove k digits.
N <= 240 and k <= N,

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Example
Given an integer A = “178542”, k = 4
return a string “12”

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解题思路

贪心的思想，从左到右扫描字符串，如果A[i] > A[i+1]则删除A[i]，直至删除了k个数字为止。这样就能使得越小的数字排在最左边，从而使得删除了k个数字后的数最小。

⚠注意两个特殊情况：

1. 如果扫描了一遍以后，已经删除了的数字(deleted_num)小于k个，甚至是A中所有数字升序排列，没有A[i] > A[i+1]的情况(此时deleted_num = 0)，则要删除末尾的(k - deleted_num)个字符。
   eg. 测试样例： A = “12345”, k = 2
2. 执行了上述算法后，还需进行后处理，删除首位的零。
   eg. 测试样例： A = “90249”, k = 2
   输出应该是“24”，而不是“024”

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代码如下：

#include <string>#include <iostream>using namespace std;class Solution {public:    /**     *@param A: A positive integer which has N digits, A is a string.     *@param k: Remove k digits.     *@return: A string     */    string DeleteDigits(string A, int k) {        // wirte your code here        int deleted_num = 0;        int n = A.size();        if(!A.empty() && n <= 240 && k < n){            for(auto i = A.begin(); i != A.end() - 1;){                if(*i > *(i + 1)){                    A.erase(i);                    deleted_num++;                    if(i != A.begin())                        i--;                    if(deleted_num == k)                        break;                }                else                    i++;            }            //⚠1            if(deleted_num < k){                A.erase(A.end() - (k - deleted_num), A.end());            }            //⚠2            auto i = A.begin();            while(*i == '0')                A.erase(i);        }        return A;    }};