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