LeetCode 402. Remove K Digits

402. Remove K Digits

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• Total Accepted: 13739
• Total Submissions: 52825
• Difficulty: Medium
• Contributors: Admin

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.

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【题目分析】

即使知道一个题目的解题思路，如何用代码简洁优美地来表达也是一种艺术。

这个题目的大意是去掉一个数中出现的若干数字，使得得到的数是可能得到的结果中最小的那个。

【思路分析】

通过观察不难发现，我们每次都去掉那个从头开始搜索的局部最大值，采用这种贪心策略可以得到最优解。在得到结果后我们还要对开始字符为0的特殊情况进行处理。但是在实现的时候，自己的代码不仅冗余，而且对边界情况处理很不到位，导致最后的结果不正确。看了讨论区大神的代码，真是很惭愧。

【java代码】

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);
    }
}