402. Remove K Digits/738.Monotone Increasing Digits/321. Create Maximum Number

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:
    //使用栈
    string removeKdigits(string num, int k)
    {
        if (k >= num.size())
            return "";
        string res = "";
        int count = k;
        for (char c : num)
        {
            while (count >  && res.size() >  && res.back() > c)
            {
                res.pop_back();
                count--;
            }
            res.push_back(c);
        }
        res = res.substr(,num.length()-k);
        while (res.empty()== false && res[] == '')
            res.erase(res.begin());
        return res.length()<= ? "" : res;
    }
    //深搜
    vector<int> all;
    string removeKdigits(string num, int k)
    {
        if (k >= num.length())
            return "";
        vector<int> bit(num.length(), );
        vector<int> flag(num.length(), );
        for (int i = ; i < bit.size(); i++)
            bit[i] = (int)(num[i] - '');
        getAll(,k, flag, bit);
        int minRes = numeric_limits<int>::max();
        for (int one : all)
            minRes = min(minRes, one);
        return to_string(minRes);
    }
    void dfs(int begin,int k,vector<int>& flag,vector<int>& bit)
    {
        if (k == )
        {
            int a = ;
            for (int i = ; i < bit.size(); i++)
            {
                if (flag[i] == )
                    a = a *  + bit[i];
            }
            all.push_back(a);
        }
        else if (begin >= flag.size())
            return;
        else
        {
            dfs(begin + , k, flag, bit);
            flag[begin] = ;
            dfs(begin + , k - , flag, bit);
            flag[begin] = ;
        }
    }
};

738.leetcode: Monotone Increasing Digits

Given a non-negative integer N, find the largest number that is less than or equal to N with monotone increasing digits.

(Recall that an integer has monotone increasing digits if and only if each pair of adjacent digits x and y satisfy x <= y.)

Example 1:

Input: N = 10
Output: 9

Example 2:

Input: N = 1234
Output: 1234

Example 3:

Input: N = 332
Output: 299

class Solution
{
public:
    int monotoneIncreasingDigits(int n)
    {
        string num = to_string(n);
        int begin = num.length();
        for (int i = num.length() - ; i >= ; i--)
        {
            if (num[i] >= num[i - ])
                continue;
            else
            {
                num[i - ]--;
                begin = i;
            }
        }
        for (int i = begin; i < num.length(); i++)
            num[i] = '';
        return stoi(num);
    }
};

321. Create Maximum Number:

Given two arrays of length m and n with digits 0-9 representing two numbers. Create the maximum number of length k <= m + nfrom digits of the two. The relative order of the digits from the same array must be preserved. Return an array of the k digits.

Note: You should try to optimize your time and space complexity.

Example 1:

Input:
nums1 = [3, 4, 6, 5]
nums2 = [9, 1, 2, 5, 8, 3]
k = 5
Output:
[9, 8, 6, 5, 3]

Example 2:

Input:
nums1 = [6, 7]
nums2 = [6, 0, 4]
k = 5
Output:
[6, 7, 6, 0, 4]

Example 3:

Input:
nums1 = [3, 9]
nums2 = [8, 9]
k = 3
Output:
[9, 8, 9]

class Solution
{
public:
    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k){
        int m = nums1.size(), n = nums2.size();
        vector<int> res;
        // i的取值范围要小心
        for (int i = max(, k - n); i <= min(k, m); ++i) {
            res = max(res, mergeVector(maxVector(nums1, i), maxVector(nums2, k - i)));
        }
        return res;
    }
    // 栈的思想，求取k个数的最大值
    vector<int> maxVector(vector<int> nums, int k) {
　　　   // 丢的方式比捡的方式实现
        int drop = nums.size() - k;
        vector<int> res;
        for (int num : nums) {
            while (drop && res.size() && res.back() < num) {
                res.pop_back();
                --drop;
            }
            res.push_back(num);
        }
        res.resize(k);
        return res;
    }
    // 和有序数组外排有区别，求最大的归并值
    vector<int> mergeVector(vector<int> nums1, vector<int> nums2) {
        vector<int> res;
        while (nums1.size() + nums2.size()) {
            vector<int> &tmp = nums1 > nums2 ? nums1 : nums2;
            res.push_back(tmp[]);
            tmp.erase(tmp.begin());
        }
        return res;
    }
};