[LeetCode] 402. Remove K Digits

Given string num representing a non-negative integer num, and an integer k, return the smallest possible integer after removing k digits from num.

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.

Constraints:

• 1 <= k <= num.length <= 105
• num consists of only digits.
• num does not have any leading zeros except for the zero itself.

移掉 K 位数字。

给你一个以字符串表示的非负整数 num 和一个整数 k ，移除这个数中的 k 位数字，使得剩下的数字最小。请你以字符串形式返回这个最小的数字。

思路是贪心，会用到单调栈。这里我们需要思考一个问题，什么样的数字才是最小的？首先高位的数字需要尽可能的低，可以尝试用 stack，从左往右把每一位 push 到 stack，当 stack 不为空且栈顶元素比要 push 进去的元素要大的时候，pop 出栈顶元素，直到丢弃了 K 个数字为止。但是这个题也有一些 corner case，比如第二个例子，10200，放弃一个数字，按照之前的思路，会丢弃掉 1，但是剩下的部分是 0200，需要处理掉开头的 0；再看第三个例子，10，需要去掉 2 个数字，会使得 res 成为 0。

时间O(n)

空间O(n)

Java实现

class Solution {
    public String removeKdigits(String num, int k) {
        // corner case
        if (k == num.length()) {
            return "0";
        }

        // normal case
        Stack<Character> stack = new Stack<>();
        for (int i = 0; i < num.length(); i++) {
            while (k > 0 && !stack.isEmpty() && stack.peek() > num.charAt(i)) {
                stack.pop();
                k--;
            }
            stack.push(num.charAt(i));
        }

        while (k > 0) {
            stack.pop();
            k--;
        }

        StringBuilder sb = new StringBuilder();
        while (!stack.isEmpty()) {
            sb.append(stack.pop());
        }
        sb.reverse();

        int index = 0;
        // skip the leading zeros
        while (index < sb.length() && sb.charAt(index) == '0') {
            index++;
        }
        return index == sb.length() ? "0" : sb.substring(index);
    }
}

 

JavaScript实现

/**
 * @param {string} num
 * @param {number} k
 * @return {string}
 */
var removeKdigits = function(num, k) {
    // corner case
    if (num === null || num.length === 0) {
        return '0';
    }

    // normal case
    let stack = [];
    for (let i = 0; i < num.length; i++) {
        while (k > 0 && stack.length && num.charAt(i) < stack[stack.length - 1]) {
            stack.pop();
            k--;
        }
        stack.push(num.charAt(i));
    }

    while (k > 0) {
        stack.pop();
        k--;
    }

    let sb = stack.join('');
    let res = 0;
    while (res < sb.length && sb.charAt(res) === '0') {
        res++;
    }
    return res === sb.length ? '0' : sb.substring(res);
};

 

LeetCode 题目总结