Leetcode:402. Remove K Digits (Week 7)

Description:

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.

解题思路及算法分析

本题是medium，比较麻烦的是要弄清楚其规律，很容易遗漏0的情况。考虑怎么解决，就要先弄清楚去数规律：比如去k个数字，要是前k个里面包含0，那么就要把0留下，最后自动去除；若没有0，要去的这个数，一定是从左到右，第一次比它后面的那个数字大的数字，但是若数字是非递减的，则去掉后面最大的数字保留第一数的数即可。酱紫就可以解决所有情况，包括是0的情况。
注意：最后输出结果的时候要去掉前面的0。

代码

class Solution {public:    string removeKdigits(string num, int k) {        int n = num.size();        if(k >= n) return "0";        while(k) {            int i = 0;            bool flag = false;            for (i = 0; i<num.size(); i++) {                if(num[i] > num[i+1]) {                    flag = true;                    num = num.erase(i,1);                    break;                }            }            if(flag == false) {                num = num.substr(0,num.size()-1);            }            k--;        }        while(num.size() > 1 && num[0] == '0') {            num = num.substr(1);        }        return num;    }};