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.
316. Remove Duplicate Letters
321. Create Maximum Number
402. Remove K Digits
这三道题都用到了 stack 来求原序列中不打乱相对次序的最小子序列的技巧:
while(k>0 && !stack.isEmpty() && stack.peek()>num.charAt(i)) 栈内元素是由大到小排列的, 站内元素也是需要去除的,
保证最高位的元素最小.所以遍历时碰到比栈内小的元素就pop
public class Solution { public String removeKdigits(String num, int k) { int len = num.length(); //corner case if(k==len) return "0"; Stack<Character> stack = new Stack<>(); int i =0; while(i<num.length()){ //whenever meet a digit which is less than the previous digit, discard the previous one while(k>0 && !stack.isEmpty() && stack.peek()>num.charAt(i)){ stack.pop(); k--; } stack.push(num.charAt(i)); i++; } // corner case like "1111" while(k>0){ stack.pop(); k--; } //construct the number from the stack StringBuilder sb = new StringBuilder(); while(!stack.isEmpty()) sb.append(stack.pop()); sb.reverse(); //remove all the 0 at the head while(sb.length()>1 && sb.charAt(0)=='0') sb.deleteCharAt(0); return sb.toString(); } }