LeetCode Week 12
402. 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.
solutions:
本題意是要求給定一個表示為字串的非負整數num,從該數字中刪除k個數字,以便新的數字是最小的。
我們能想到的是給定的字串不能改變相對位置,只能刪除其中的數字。所以很容易知道可以從最高位開始。比如6位num, k = 2.那麼刪除後剩下4位,那麼最高位肯定是在前三位中取(尤其地若在第三位取最高位,則字串就確定了)
string removeKdigits(string num, int k) { string res; helper(num, k, res); return res; } void helper(string num, int k, string& res) { if (k == num.size()) { if (res.size() == 0) res = "0"; return; } if (k == 0) { res += num; if (res.size() == 0) res = "0"; return; } int minid = 0; for (int i = 0; i <= k; i++) { if (num[i] < num[minid]) minid = i; } if (res.size() != 0 || num[minid] != '0') res += num[minid]; helper(num.substr(minid+1, num.size()-minid-1), k-minid, res); return; }