Java Greedy Solution Accepted

• This is a recursive solution. First, find the minimum digit within range[0, k]. determine how many digit to delete in order to make it the first digit. Retain that digit, recursive find the minimum number behind it.

``````public class Solution {
public String removeKdigits(String num, int k) {
String result = helper(num, k);
int i = 0;
while (i < result.length()-1) {
if (result.charAt(i) != '0') break;
i++;
}
result = result.substring(i, result.length());
if (result.length() == 0) return "0";
return result;
}

private String helper(String num, int k) {
if (num.length() <= k) return "";
if (k == 0) return num;
char min = num.charAt(0);
int minIndex = 0;
for (int i = 1; i <= k; i++) {
if (num.charAt(i) < min) {
min = num.charAt(i);
minIndex = i;
}
}
String num2 = helper(num.substring(minIndex+1, num.length()), k-minIndex);
return num.charAt(minIndex)+num2;
}
}``````

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• Could you explain your code more? Why it works?

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