Description
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.
Solution
Greedy & Stack, time O(n), space O(n)
假設(shè)num的組成是"abc....."是辕，如果刪掉'a'吏夯，那么剩下的是"bc...."，一定是由b打頭术奖；如果刪掉任何a之后的字符站楚，剩下的string必定由a打頭脱惰。于是比較a和b的大小即可，用Greedy的思路選擇最優(yōu)解即可窿春。
注意需要考慮a = b的情況拉一，在這種情況下需要先遍歷到后面的元素才知道怎么處理a和b（考慮"223"和"221"），所以用Stack來(lái)實(shí)現(xiàn)就顯得很合理了旧乞。
class Solution {
public String removeKdigits(String num, int k) {
if ("0".equals(num) || k < 1) {
return num;
}
Stack<Character> stack = new Stack<>();
for (char c : num.toCharArray()) {
// whenever meet a digit which is less than the previous digit,
// discard the previous one to keep chars in stack are in asc order
while (k > 0 && !stack.empty() && stack.peek() > c) {
stack.pop();
--k;
}
if (stack.empty() && c == '0') { // discard '0' at the head
continue;
}
stack.push(c);
}
while (!stack.empty() && k-- > 0) { // corner case "1234", 2
stack.pop();
}
StringBuilder sb = new StringBuilder();
while (!stack.empty()) {
sb.insert(0, stack.pop()); // construct number
}
return sb.length() == 0 ? "0" : sb.toString();
}
}
