402. Remove K Digits

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.

class Solution {
    public String removeKdigits(String num, int k) {
        int le = num.length();
        if(k >= le || le == 0) return "0";
        while(true){
            le = num.length();
            if(le == 0) return "0";
            if(k-- == 0) return num;
            if(num.charAt(1) == '0'){
                int zeroind = 1;
                while(zeroind < le && num.charAt(zeroind) == '0'){
                    zeroind++;
                }
                num = num.substring(zeroind);
            }
            else{
                int i = 0;
                while(i < le-1){
                    if(num.charAt(i) > num.charAt(i+1)) 
                    {num = num.substring(0, i) + num.substring(i+1);
                    break;
                    }
                    else i++;
                }
                if(i == le-1) num = num.substring(0, i);
            }
        }
    }
}

参考 https://www.cnblogs.com/271934Liao/p/7083544.html

有几个注意点

1. 每次删除后num位数会减小，需要重新计算num长度

2. 分两种情况 

　　A。第二位是0，那就删掉第一位和第二位以及后面连续的0直到没0，注意这只算删了一位。如果删完了就直接返回0

　　B。第二位不是0，向后检查直到发现第一个i>i+1为止，为此要设置i<length-1，注意一次只删一位，删完立刻break。如果删到最后一位还是递增，直接删除最后一位

 

class Solution {
    public String removeKdigits(String num, int k) {
        int n = num.length();
        if(n == k) return "0";
        
        Stack<Character> stack = new Stack();
        int i = 0;
        while(i < n) {
            while(k > 0 && !stack.isEmpty() &&  stack.peek() > num.charAt(i)) {
                k--;
                stack.pop();
            }
            stack.push(num.charAt(i));
            i++;
        }
        while(k != 0) {
            k--;
            stack.pop();
        }
        
        StringBuilder sb = new StringBuilder();
        while(!stack.isEmpty()) sb.append(stack.pop());
        
        while(sb.length() > 1 && sb.charAt(sb.length() - 1) == '0') sb.deleteCharAt(sb.length() - 1);
        
        return sb.reverse().toString();
    }
}

神说，要用stack，本质上是stack + greedy的方法

greedy，把所有char存到stack中，如果stack peek 大于当前的char，就要把这个stack peek给pop掉。如果stack push完所有的char了k还没用完，说明有相等的数或者升序1234出现，那么继续pop完所有的。

然后把所有的0开头的去掉剩下的就是答案。