标签：code ++ leading string 入栈 push put 递增 with
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.
思路：其基本思想是利用栈尽量维持一个递增的序列，也就是说将字符串中字符依次入栈，如果当前字符串比栈顶元素小，并且还可以继续删除元素，那么就将栈顶元素删掉，这样可以保证将当前元素加进去一定可以得到一个较小的序列．也可以算是一个贪心思想．最后我们只取前len-k个元素构成一个序列即可，如果这样得到的是一个空串那就手动返回０．还有一个需要注意的是字符串首字符不为０
class Solution { public String removeKdigits(String num, int k) { if(num == null || num.length() == 0){ return null; } Stack<Integer> stack = new Stack<>(); for(int i = 0; i<num.length(); i++){ int cur = num.charAt(i) - ‘0‘; while(!stack.isEmpty() && cur < stack.peek() && num.length() - i - 1 >= num.length()-k-stack.size()){ stack.pop(); } if(stack.size() < num.length()-k){ stack.push(cur); } } StringBuilder res = new StringBuilder(); int count = 0; while (!stack.isEmpty()){ res.insert(0, stack.pop()); } while (res.length() > 0 && res.charAt(0) == ‘0‘){ res.deleteCharAt(0); } if(res.length() == 0){ return "0"; } return res.toString(); } }
LeetCode-Microsoft-Remove K Digits
标签：code ++ leading string 入栈 push put 递增 with
原文地址：https://www.cnblogs.com/incrediblechangshuo/p/8970525.html
