0402._Remove_K_Digits

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402. Remove K Digits

难度: Medium

刷题内容

原题连接

内容描述

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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.

解题方案

思路 1
**- 时间复杂度: O(N)**- 空间复杂度: O(N)**

  • 使用栈的思路
  • 如果n是num的长度，我们要去除k个，那么需要剩下n-k个数，定义一个result数组用于保存剩下的字符，与result中最后一个字符相比，比它小，
  • 栈中最后一个字符出栈，该字符进栈，否则该字符直接进栈。值得注意的是在删除k个数之后，若剩下的数前面有0，应该去掉。

代码：

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/**
 * @param {string} num
 * @param {number} k
 * @return {string}
 */
var removeKdigits = function(num, k) {
    let stack = [], numDigits = num.length;
    for (let i = 0; i < numDigits; i++) {
        while(k > 0 && stack.length && stack[stack.length - 1] > num[i]) {
            stack.pop();
            k--;
        }
        stack.push(num[i]);
    }
    stack = k > 0 ? stack.slice(0, -k) : stack;
    return stack.join('').replace(/^0+/, '') || '0';
};