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.
explain
題意是從該字符串中刪除K位字符，使得結果是最小的。一般看到最小的，如果不想暴力的做的話，要用貪婪的策略。我這里給到的方法，就是把前一個比后一個大的數刪除，這樣一趟下來，位比較前，又比較大的數會被刪掉。如果一趟遍歷，刪除的位數不足，觀察樣例，是要把從后面把剩余的K位剔除。剔除后，如果字符串的第一位是0，還得繼續剔除首位0的情況。因此，整個答案就出來。
solution
class Solution {
public:
string removeKdigits(string num, int k) {
int len = num.length();
if (len <= k) return "0";
string res = "";
for (int i = 0; i < num.length(); i++) {
while (k != 0 && !res.empty() && num[i] < res.back()) {
res.pop_back();
k--;
}
res.push_back(num[i]);
}
if (k) res = res.substr(0, res.length() - k);
while(!res.empty() && res[0] == '0') res.erase(res.begin());
if (res.empty()) return "0";
else return res;
}
};
