题目链接
tag:
- Medium；
- Greedy；
question：
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：
这道题让我们从给定的数字去掉k位，要使得留下来的数字最小；其基本思想是利用stack维护一个递增的序列，也就是说将字符串中的字符依次入栈，如果当前字符串比栈顶元素小，并且还可以继续删除元素，那么就将栈顶元素去掉，这样就可以保证将当前元素加进去一定可以得到一个较小的序列，算做一个Greedy的思想，最后我们只取前len-k个元素构成一个序列即可，如果这样的到的是一个空串那就手动返回0，当然需要注意字符串首字符不为0。AC代码如下：
class Solution {
public:
string removeKdigits(string num, int k) {
string res = "";
int n = num.size(), keep = n-k;
for (char c : num) {
while (k && num.size() && res.back() > c) {
res.pop_back();
--k;
}
res.push_back(c);
}
res.resize(keep);
while (!res.empty() && res[0] == '0') {
res.erase(res.begin());
}
return res.empty() ? "0" : res;
}
};
