leetcode——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.

解析：若k=num.length,則返回0。否則的話，利用貪心算法，從字符串中刪除k個數字，則將留下num.length-k個數字。顯然，留下的第一個數字肯定位於字符串的前k+1位，爲了讓新數字儘可能小，所以最高位要儘可能小，故選擇第一次出現的最小的數字作爲最高位，並從num中刪去；同理，留下的第二個數字位於當前num中前k+1位...極端情況下算法時間複雜度爲O(n^2)。代碼如下：

class Solution {
public:
	string removeKdigits(string num, int k) {
		if(k == num.length())	return "0";
		int p = 0, l=num.length()-k;
		string ans;
		for(int i=0; i<l; i++)
		{
			char min = num[p];
			for(int j=p+1; j<=k; j++)
				if(num[j] < min)
				{
					min = num[j];
					p = j;
				}
			ans += num[p];
			num.erase(p, 1);
		}
		while(ans.length() > 1 && ans[0] == '0')	ans.erase(0, 1);
		return ans;
	}
};

不過呢，算法還可以適當優化，例如，由於p只增不減，所以當每次選擇的的數字都位於第k位時，可以不需要繼續搜索下去，直接將後面的子串copy即可，可以適當增大時間效率。代碼略作修改：

class Solution {
public:
	string removeKdigits(string num, int k) {
		if(k == num.length())	return "0";
		int p = 0, l=num.length()-k;
		string ans;
		for(int i=0; i<l; i++)
		{
			char min = num[p];
			for(int j=p+1; j<=k; j++)
				if(num[j] < min)
				{
					min = num[j];
					p = j;
				}
			if(p == k)
			{
				ans += num.substr(p, num.length()-k);
				break;
			}
			ans += num[p];
			num.erase(p, 1);
		}
		while(ans.length() > 1 && ans[0] == '0')	ans.erase(0, 1);
		return ans;
	}
};