402-删除k位使得最小

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.

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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.

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问题描述

给定以字符串表示的正整数num，从num中去除k位使得新得到的num最小。

注意

• num的长度大于等于k，小于10002
• num没有前零

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问题分析

用stk数组来保存处理结果，k来保存可以删除的位数。
若当前元素比stk中的元素小，那么删除stk中的元素可以获得更小的数

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解法

public class Solution {
    public String removeKdigits(String num, int k) {
        int digits = num.length() - k;
        char[] stk = new char[num.length()];
        int top = 0;

        // k keeps track of how many characters we can remove
        // if the previous character in stk is larger than the current one
        // then removing it will get a smaller number
        // but we can only do so when k is larger than 0
        for (int i = 0; i < num.length(); ++i) {
            char c = num.charAt(i);
            while (top > 0 && stk[top-1] > c && k > 0) {
                top -= 1;
                k -= 1;
            }
            stk[top++] = c;
        }

        // find the index of first non-zero digit
        int idx = 0;
        while (idx < digits && stk[idx] == '0') idx++;

        return idx == digits? "0": new String(stk, idx, digits - idx);
    }
}