402 Remove K Digits
402 Remove K Digits
1. 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.
2. Implementation
思路:这题思路和Remove Duplicate Letters一样,利用单调栈保持输出结果的单调性,只是要注意一些Corner cases
class Solution {
public String removeKdigits(String num, int k) {
int len = num.length();
if (k == len) {
return "0";
}
Stack<Character> stack = new Stack<>();
for (char digit : num.toCharArray()) {
while (k > 0 && !stack.isEmpty() && stack.peek() > digit) {
stack.pop();
--k;
}
stack.push(digit);
}
while (k > 0) {
stack.pop();
--k;
}
StringBuilder res = new StringBuilder();
while (!stack.isEmpty()) {
res.append(stack.pop());
}
while (res.length() > 1 && res.charAt(res.length() - 1) == '0') {
res.deleteCharAt(res.length() - 1);
}
return res.reverse().toString();
}
}
3. Time & Space Complexity
时间和空间复杂度都是O(n)
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