题目链接: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:
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:其基本思想是利用栈尽量维持一个递增的序列,也就是说将字符串中字符依次入栈,如果当前字符串比栈顶元素小,并且还可以继续删除元素,那么就将栈顶元素删掉,这样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.也可以算是一个贪心思想.最后我们只取前len-k个元素构成一个序列即可,如果这样得到的是一个空串那就手动返回0.还有一个需要注意的是字符串首字符不为0
代码如下:
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;
}
};