[LeetCode] 402. Remove K Digits

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.

移掉K位数字。

题意是给一个以字符串表示的数字和一个数字K，请你从字符串中移除K位数字，使得剩下的部分组成的数字是最小的。

思路是贪心，会用到单调栈。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，所以想到用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 res = 0;
        // skip the leading zeros
        while (res < sb.length() && sb.charAt(res) == '0') {
            res++;
        }
        return res == sb.length() ? "0" : sb.substring(res);
    }
}

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