Leetcode: 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:
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.

Greedy + Stack: 用一个栈维护最后要留存下来的digits

需要注意的是：如果遍历到string的某个char, string后面的char数刚刚好能填满这个栈，那么即使这个char比栈顶还要小，也不出栈，直接入栈

最后要删除leading 0

 1 public class Solution {
 2     public String removeKdigits(String num, int k) {
 3         if (num.length()==0 || k>=num.length()) return "0";
 4         char[] arr = num.toCharArray();
 5         Stack<Character> stack = new Stack<Character>();
 6         StringBuilder res = new StringBuilder();
 7         int size = arr.length - k;
 8         for (int i=0; i<arr.length; i++) {
 9             while (!stack.isEmpty() && arr[i]<stack.peek() && (size-stack.size()+1 <= arr.length-i)) {
10                 stack.pop();
11             }
12             if (size > stack.size())
13                 stack.push(arr[i]);
14         }
15         while (!stack.isEmpty()) {
16             res.insert(0, stack.pop());
17         }
18         while (res.length()>1 && res.charAt(0)==‘0‘) res.deleteCharAt(0);
19         return res.toString();
20     }
21 }

用char[]实现栈，思路一样，要快很多，beat96%

 1 public class Solution {
 2     public String removeKdigits(String num, int k) {
 3         int remain = num.length() - k;
 4         char[] numArray = num.toCharArray(), res = new char[remain];
 5         int index = 0;
 6         for(int i = 0; i < numArray.length; i++) {
 7             while((numArray.length - i > remain - index) && (index > 0 && numArray[i] < res[index - 1])) index--;
 8             if(index < remain) res[index++] = numArray[i];
 9         }
10
11         // check leading zeroes
12         index = -1;
13         while(++index < remain) {
14             if(res[index] != ‘0‘) break;
15         }
16         String s = new String(res).substring(index);
17
18         return s.length() == 0 ? "0" : s;
19     }
20 }