问题:
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 = 3Output: "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 = 1Output: "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 = 2Output: "0"Explanation: Remove all the digits from the number and it is left with nothing which is 0.
解决:
①
有一个N为长的数字,用字符串来代替了,现在要求你将它删除K位,使得其得到的结果最小。对于每次删除的情况有两种
(1)假如第一个数不为0,第二个数为0,那么我们删除第一个数,就相当于数量级减少2,这样比删除得到的数其他任何一个方法都小
(2)另一种情况,我们找到第一次遍历的局部最大值,即遍历num第一个满足num.charAt(i)>num.charAt(i+1)的值,删除这个点,得到的值最小。这里就是贪心算法,每次删除一个局部最大。
class Solution {//6ms
public String removeKdigits(String num, int k) { StringBuilder sb = new StringBuilder(); int len = num.length(); char[] stack = new char[len]; int count = 0; for (int i = 0;i < len;i ++){ while(count != 0 && k > 0 && num.charAt(i) < stack[count - 1]){//根据贪心算法删除比后一个大的值 count --; k --; } stack[count ++] = num.charAt(i); } int start = 0; while(start < count && stack[start] == '0') start ++; return start >= count - k ? "0" : new String(stack,start,count - start - k); } }