[leetcode] 402. Remove K Digits 解題報告

題目鏈接：https://leetcode.com/problems/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.

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：其基本思想是利用棧儘量維持一個遞增的序列，也就是說將字符串中字符依次入棧，如果當前字符串比棧頂元素小，並且還可以繼續刪除元素，那麼就將棧頂元素刪掉，這樣可以保證將當前元素加進去一定可以得到一個較小的序列．也可以算是一個貪心思想．最後我們只取前len-k個元素構成一個序列即可，如果這樣得到的是一個空串那就手動返回０．還有一個需要注意的是字符串首字符不爲０

代碼如下：

class Solution {
public:
    string removeKdigits(string num, int k) {
        string ans;
        int n = k, len = num.size(), cnt = 0;
        for(auto val: num)
        {
            while(!ans.empty() && n > 0 && val < ans.back())
            {
                n--;
                ans.pop_back();
            }
            ans.push_back(val);
        }
        while(ans[cnt]=='0') cnt++;
        ans = ans.substr(cnt, len-k-cnt);
        return !ans.size()?"0":ans;
    }
};