[LeetCode] 402. Remove K Digits Java

题目：

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.

 题意及分析：有一个N为长的数字，用字符串来代替了，现在要求你将它删除K位，使得其得到的结果最小。对于每次删除的情况有两种

(1)假如第一个数不为0，第二个数为0，那么我们删除第一个数，就相当于数量级减少2，这样比删除得到的数其他任何一个方法都小

(2)另一种情况，我们找到第一次遍历的局部最大值，即遍历num第一个满足num.charAt(i)>num.charAt(i+1)的值，删除这个点，得到的值最小。这里就是贪心算法，每次删除一个局部最大

代码：

public class Solution {
    public String removeKdigits(String num, int k) {
        int n=0;
        while(true){
        	n =num.length();
        	if(n <= k || n == 0) return "0";
        	if(k-- == 0){
        // 		System.out.println(num);
        		return num;
        	};
        	if(num.charAt(1)=='0'){		//第二位数为0，则删除前面一位数
        		int firstNotZero = 1;
        		while(firstNotZero<n&&num.charAt(firstNotZero)=='0') firstNotZero++;
        		num=num.substring(firstNotZero);
        	}else{		//不然寻找第一个下降的数
        		int i=0;
        		while(i<n-1){
        			if(num.charAt(i)>num.charAt(i+1)){
        				num=num.substring(0, i)+num.substring(i+1);
        				break;
        			}else
        				i++;
        		}
        		if(i==n-1)  num=num.substring(0, i);
        	}
        }
    }
}

　　

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