題目：

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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.

?題意及分析：有一個N為長的數字，用字符串來代替了，現在要求你將它刪除K位，使得其得到的結果最小。對于每次刪除的情況有兩種

(1)假如第一個數不為0，第二個數為0，那么我們刪除第一個數，就相當于數量級減少2，這樣比刪除得到的數其他任何一個方法都小

(2)另一種情況，我們找到第一次遍歷的局部最大值，即遍歷num第一個滿足num.charAt(i)>num.charAt(i+1)的值，刪除這個點，得到的值最小。這里就是貪心算法，每次刪除一個局部最大

代碼：

public class Solution {public String removeKdigits(String num, int k) {int n=0;while(true){n =num.length();if(n <= k || n == 0) return "0";if(k-- == 0){// System.out.println(num);return num;};if(num.charAt(1)=='0'){ //第二位數為0，則刪除前面一位數int firstNotZero = 1;while(firstNotZero<n&&num.charAt(firstNotZero)=='0') firstNotZero++;num=num.substring(firstNotZero);}else{ //不然尋找第一個下降的數int i=0;while(i<n-1){if(num.charAt(i)>num.charAt(i+1)){num=num.substring(0, i)+num.substring(i+1);break;}elsei++;}if(i==n-1) num=num.substring(0, i);}}} }

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Seen this question in a real interview before??

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轉載于:https://www.cnblogs.com/271934Liao/p/7083544.html

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