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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:
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The length of num is less than 10002 and will be ≥ k.
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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.思路分析
利用单调栈的思路，将前面比当前字符大的字符都删除即可。
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-08-16 19:48:26
*/
public String removeKdigits(String num, int k) {
if (num == null || num.isEmpty()) {
return num;
}
if (num.length() <= k) {
return "0";
}
char[] chars = num.toCharArray();
StringBuilder stack = new StringBuilder();
stack.append(chars[0]);
for (int i = 1; i < chars.length; i++) {
char c = chars[i];
while (!stack.isEmpty() && stack.charAt(stack.length() - 1) > c && k > 0) {
stack.deleteCharAt(stack.length() - 1);
k--;
}
stack.append(c);
}
for (int i = k; i > 0; i--) {
stack.deleteCharAt(stack.length() - 1);
}
while (!stack.isEmpty() && stack.charAt(0) == '0') {
stack.deleteCharAt(0);
}
return stack.isEmpty() ? "0" : stack.toString();
}