标签：desc ++ rem har pre 元素 class ack int
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:
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.
题目含义：给定的整数中删除k位后尽可能得到最小值
1 public String removeKdigits(String num, int k) { 2 Stack<Character> digs = new Stack<>(); 3 int length = num.length(); 4 if (k == 0) return num; 5 if (k == length) return "0"; 6 int i = 0; 7 while (i < length) { 8 9 while (k > 0 && !digs.isEmpty() && num.charAt(i) < digs.peek()) { 10 //保证顶上的元素小于等于num.charAt(i) 11 digs.pop(); 12 k--; //相当于删除一位较大的值 13 } 14 digs.push(num.charAt(i));//保证栈从顶到低的值是递减的 15 i++; 16 } 17 while (k > 0) { //k为数字还没有删除够，继续删除 18 digs.pop(); 19 k--; 20 } 21 StringBuilder sb = new StringBuilder(); 22 while (!digs.isEmpty()) { 23 sb.append(digs.pop()); //构成由高到底的字符串 24 } 25 sb.reverse();//翻转成由底到高的字符串，因为在push的时候是按照i顺序push的，所以反转后的字符串中每个字符的先后顺序和原来保持一致 26 while (sb.length() > 1 && sb.charAt(0) == ‘0‘) { 27 sb.deleteCharAt(0); 28 } 29 return sb.toString(); 30 }
标签：desc ++ rem har pre 元素 class ack int
原文地址：http://www.cnblogs.com/wzj4858/p/7726044.html
