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;

    }

};

巴特西

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

最新文章

热门文章