Remove K Digits

Posted 2020-10-04

题目：

　　

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.

public String removeKdigits(String num, int k) {
　　/**
　　*用StringBuffer当做栈
　　*向栈中加入元素，如果新加入的元素小于栈顶元素弹出
　　*/
　　StringBuffer result = new StringBuffer();
　　int count = k, i;
　　for (i = 0; i < num.length(); ++i) {
　　　　while (result.length() > 0 && count > 0 && num.charAt(i) < result.charAt(result.length() - 1)) {
　　　　　　result.deleteCharAt(result.length() - 1);
　　　　　　count--;
　　　　}
　　　　result.append(num.charAt(i));
　　}
　　for (i=0;i<result.length();++i) { //删除结果中开始多余的0
　　　　if (result.charAt(i) != ‘0‘) {
　　　　　　break;
　　　　}
　　}
　　//计算结果字符串 //如果结果中全是0，防止数组过界
　　String re = result.substring(Math.min(i,result.length()-count), result.length() - count);
　　return re.equalsIgnoreCase("") ? "0" : re; //处理结果为空的情况
}