LeetCode-Microsoft-Remove K Digits

最新推荐文章于 2018-11-13 07:45:50 发布

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最新推荐文章于 2018-11-13 07:45:50 发布

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原文链接：http://www.cnblogs.com/incrediblechangshuo/p/8970525.html

本文介绍了一种算法，该算法通过移除指定数量的数字来找到由剩余数字组成的最小可能数值。采用类似栈的数据结构维持递增序列，确保每次添加的新元素都能构成最小序列。

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.

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();
    }
}