leetcode [402]Remove K Digits

最新推荐文章于 2025-08-15 16:44:13 发布

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最新推荐文章于 2025-08-15 16:44:13 发布

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文章标签：数据结构与算法 java

原文链接：http://www.cnblogs.com/xiaobaituyun/p/11027513.html

本文探讨了一种算法，旨在从给定的非负整数字符串中删除k个数字，以获得可能的最小数值。通过使用栈结构，文章详细解释了如何在删除过程中保持数字的最小化，同时避免前导零的问题。

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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.

题目大意：

给定一个非负的整数字符串，从这个字符串中删除k个数字，得到最大的一个字符串。

解法：

利用一个stack存储字符，一旦栈顶元素大于遍历的元素，便将栈顶元素去除。然后将所有栈顶元素进行拼接。如果说删除的元素还没有k个，这时从栈底到栈顶都是递增的，直接删除栈顶元素即可。还要对元素进行处理，防止有前导0元素的出现。

java:

class Solution {
    public String removeKdigits(String num, int k) {
        Stack<Character>stack=new Stack<>();
        stack.push(num.charAt(0));
        int i=1;
        while (i<num.length()){
            while (!stack.empty() && stack.peek()>num.charAt(i) && k>0){
                stack.pop();
                k--;
            }
            stack.push(num.charAt(i));
            i++;
        }
        while (k>0){
            stack.pop();
            k--;
        }
        StringBuilder sb=new StringBuilder();
        for (Character c:stack){
            sb.append(c);
        }
        while (sb.length()>0 && sb.charAt(0)=='0'){
            sb.deleteCharAt(0);
        }

        return sb.length()==0?"0":sb.toString();
    }
}

转载于:https://www.cnblogs.com/xiaobaituyun/p/11027513.html