Wednesday, September 21, 2016

LeetCode 402 - Remove K Digits


https://leetcode.com/problems/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.
X. Using Ordered Stack
Use StringBuilder as stack
https://discuss.leetcode.com/topic/59580/short-10-lines-o-n-java-code
    public static String removeKdigits(String num, int k) {
        StringBuilder sb = new StringBuilder();
        for(char c : num.toCharArray()) {
            while(k > 0 && sb.length() != 0 && sb.charAt(sb.length() - 1) > c) {
                sb.setLength(sb.length() - 1);
                k--;
            }
            if(sb.length() != 0 || c != '0') sb.append(c);  // Only append when it is not leading zero
        }
        if(k >= sb.length()) return "0";
        sb.setLength(sb.length() - k);  // use all remaining k
        return sb.toString();  
    }
setLength(len -1) O(1)
public void setLength(int newLength) {
    if (newLength < 0)
        throw new StringIndexOutOfBoundsException(newLength);    ensureCapacityInternal(newLength);
    if (count < newLength) {
        Arrays.fill(value, count, newLength, '\0');    }

    count = newLength;}
Use array as stack
https://discuss.leetcode.com/topic/59412/a-greedy-method-using-stack-o-n-time-and-o-n-space
https://discuss.leetcode.com/topic/59501/6ms-java-solution-with-detailed-comment
    public String removeKdigits(String num, int k) {
        int digits = num.length() - k;
        char[] stk = new char[num.length()];
        int top = 0;
        // k keeps track of how many characters we can remove
        // if the previous character in stk is larger than the current one
        // then removing it will get a smaller number
        // but we can only do so when k is larger than 0
        for (int i = 0; i < num.length(); ++i) {
            char c = num.charAt(i);
            while (top > 0 && stk[top-1] > c && k > 0) {
                top -= 1;
                k -= 1;
            }
            stk[top++] = c;
        }
        // find the index of first non-zero digit
        int idx = 0;
        while (idx < digits && stk[idx] == '0') idx++;
        return idx == digits? "0": new String(stk, idx, digits - idx);
    }
http://www.cnblogs.com/grandyang/p/5883736.html
http://blog.csdn.net/qq508618087/article/details/52584133
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
使得栈中的数字尽可能保持递增顺序。
http://www.voidcn.com/blog/yeqiuzs/article/p-6204756.html
    public String removeKdigits(String num, int k) {
        int len = num.length();
        //corner case
        if(k==len)        
            return "0";
            
        Stack<Character> stack = new Stack<>();
        int i =0;
        while(i<num.length()){
            //whenever meet a digit which is less than the previous digit, discard the previous one
            while(k>0 && !stack.isEmpty() && stack.peek()>num.charAt(i)){
                stack.pop();
                k--;
            }
            stack.push(num.charAt(i));
            i++;
        }
        
        // corner case like "1111"
        while(k>0){
            stack.pop();
            k--;            
        }
        
        //construct the number from the stack
        StringBuilder sb = new StringBuilder();
        while(!stack.isEmpty())
            sb.append(stack.pop());
        sb.reverse();
        
        //remove all the 0 at the head
        while(sb.length()>1 && sb.charAt(0)=='0')
            sb.deleteCharAt(0);
        return sb.toString();
    }
https://discuss.leetcode.com/topic/59646/straightforward-java-solution-using-stack
public String removeKdigits(String num, int k) {
        Stack<Integer> stack = new Stack<Integer>();
        if (num.length() == 0 || num.length() <= k)
            return "0";

        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();
        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();
    }
http://www.cnblogs.com/grandyang/p/5883736.html
这道题让我们将给定的数字去掉k位,要使得留下来的数字最小,这题跟LeetCode上之前那道Create Maximum Number有些类似,可以借鉴其中的思路,如果n是num的长度,我们要去除k个,那么需要剩下n-k个,我们开始遍历给定数字num的每一位,对于当前遍历到的数字c,进行如下while循环,如果res不为空,且k大于0,且res的最后一位大于c,那么我们应该将res的最后一位移去,且k自减1。当跳出while循环后,我们将c加入res中,最后我们将res的大小重设为n-k。根据题目中的描述,可能会出现"0200"这样不符合要求的情况,所以我们用一个while循环来去掉前面的所有0,然后返回时判断是否为空,为空则返回“0”
    string removeKdigits(string num, int k) {
        string res = "";
        int n = num.size(), keep = n - k;
        for (char c : num) {
            while (k && res.size() && res.back() > c) {
                res.pop_back();
                --k;
            }
            res.push_back(c);
        }
        res.resize(keep);
        while (!res.empty() && res[0] == '0') res.erase(res.begin());
        return res.empty() ? "0" : res;
    }
http://blog.csdn.net/mebiuw/article/details/52576884
* 这是一个非常简单的问题,贪心解决法 * 即 removeKdigits(num,k) = removeKdigits(removeKdigits(num,1),k-1) * 进行最多K轮的删除,每次从头开始寻找一位删除: * 1、要么第二位是0,这样相当于至少删了两位,很值得,必须这么做 * 2、不然,找到第一个出现下降转折的位置 删除 * */ public String removeKdigits(String num, int k) { int n; while(true){ n = num.length(); if(n <= k || n == 0) return "0"; if(k-- == 0) return num; if(num.charAt(1) == '0'){ int firstNotZero = 1; while(firstNotZero < num.length() && num.charAt(firstNotZero) == '0') firstNotZero ++; num=num.substring(firstNotZero); } else{ int startIndex = 0; while(startIndex < num.length() - 1 && num.charAt(startIndex) <= num.charAt(startIndex + 1)) startIndex ++; num=num.substring(0,startIndex)+num.substring(startIndex+1); } } }

X.
http://bookshadow.com/weblog/2016/09/18/leetcode-remove-k-digits/
解法II 问题可以转化为对偶问题:从字符串num中选取size - k个字符,使得选出的数字最小化。
首先利用字典numd保存数字'0' - '9'在原始字符串num中的出现位置。
然后循环size - k次:
按照'0' - '9'的顺序从numd中选取位置合法的数字
位置合法包含2条原则:1) 候选数字之后剩余的数字要足够多,2) 候选数字要处在已选出的数字之后
def removeKdigits(self, num, k): """ :type num: str :type k: int :rtype: str """ size = len(num) ans = '0' numd = collections.defaultdict(collections.deque) for i, n in enumerate(num): numd[n].append(i) p = 0 for x in range(size - k): for y in '0123456789': while numd[y] and numd[y][0] < p: numd[y].popleft() if numd[y] and numd[y][0] <= k + x: p = numd[y][0] ans += y numd[y].popleft() break return str(int(ans))

X.  https://discuss.leetcode.com/topic/59871/two-algorithms-with-detailed-explaination
The first algorithm is straight-forward. Let's think about the simplest case: how to remove 1 digit from the number so that the new number is the smallest possible? Well, one can simply scan from left to right, and remove the first "peak" digit; the peak digit is larger than its right neighbor. One can repeat this procedure k times, and obtain the first algorithm:
string removeKdigits(string num, int k) {
        while (k > 0) {
            int n = num.size();
            int i = 0;
            while (i+1<n && num[i]<=num[i+1])  i++;
            num.erase(i, 1);
            k--;
        }
        // trim leading zeros
        int s = 0;
        while (s<(int)num.size()-1 && num[s]=='0')  s++;
        num.erase(0, s);
        
        return num=="" ? "0" : num;
    }

http://www.1point3acres.com/bbs/thread-202732-1-1.html
给一个string, 找出lexical order 最小的, size==k的, subsequence, (note, not substring)
String findMin(String s, k){} 
e.g.
input
s=pineapple, k==3, 

output: ale
ale is the lexical order smallest subsequnce of length 3. 
我是暴力求解的: 
1. find the first occur position of distinct char. 
2. then start from that position. 
3. dfs to find lenght==3, subsequence(dfs, combination way); . visit 1point3acres.com for more.
4. find the one with smallest lexical order. 

我就是说了个大概.
pop的时候需要看后面还剩几个元素了
元素不够的时候就含泪不pop了,直接push进去
比如你说的例子,其实是
f->e->d->c->cb->cba
  1. public class Solution {
  2.         鏉ユ簮涓€浜�.涓夊垎鍦拌鍧�. 
  3.         public String removeKdigits(String num, int k) {. 1point3acres.com/bbs
  4.         if(num == null || num.length() == 0) return num;
  5.         int len = num.length();. 1point3acres.com/bbs
  6.         if(len == k) return "";
  7.         Stack<Character> stack = new Stack<Character>();
  8.         char []ch = num.toCharArray();
  9.         int i = 0, n = num.length();
  10.         while(i < len){

  11.             while(!stack.isEmpty() && stack.size() + n-i > k &&  stack.peek() > ch[i]){
  12.                 stack.pop();
  13.             }.鐣欏璁哄潧-涓€浜�-涓夊垎鍦�
  14.             stack.push(ch[i++]);
  15.         }. more info on 1point3acres.com
  16.         
  17.         // handle corner case 1111 or 1234, when k = 2. 涓€浜�-涓夊垎-鍦帮紝鐙鍙戝竷
  18.         while(stack.size() > k){-google 1point3acres
  19.             stack.pop();
  20.         }
  21.         
  22.         StringBuilder sb = new StringBuilder();
  23.         while(!stack.isEmpty()). Waral 鍗氬鏈夋洿澶氭枃绔�,
  24.           sb.insert(0, stack.pop());
  25.         
  26.         return sb.toString();
  27.         
  28.     }

  29.         public static void main(String[] args) {.鏈枃鍘熷垱鑷�1point3acres璁哄潧
  30.                 Solution s = new Solution();
  31.                 String str="xyzabc";
  32.                 int k=3;
  33.                 System.out.println(s.removeKdigits(str, k));
  34.         }. 鐗涗汉浜戦泦,涓€浜╀笁鍒嗗湴
  35. }

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