[LintCode] Create Maximum Number 创建最大数

Posted 2020-07-26 Grandyang

 

Given two arrays of length m and n with digits 0-9 representing two numbers. Create the maximum number of length k <= m + n from digits of the two. The relative order of the digits from the same array must be preserved. Return an array of the k digits. You should try to optimize your time and space complexity.
Have you met this question in a real interview?
Example

Given nums1 = [3, 4, 6, 5], nums2 = [9, 1, 2, 5, 8, 3], k = 5
return [9, 8, 6, 5, 3]

Given nums1 = [6, 7], nums2 = [6, 0, 4], k = 5
return [6, 7, 6, 0, 4]

Given nums1 = [3, 9], nums2 = [8, 9], k = 3
return [9, 8, 9]

 

LeetCode上的原题，请参见我之前的博客Create Maximum Number。

 

class Solution {
public:
    /**
     * @param nums1 an integer array of length m with digits 0-9
     * @param nums2 an integer array of length n with digits 0-9
     * @param k an integer and k <= m + n
     * @return an integer array
     */
    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {
        vector<int> res;
        int m = nums1.size(), n = nums2.size();
        for (int i = max(0, k - n); i <= min(k, m); ++i) {
            res = max(res, mergeVec(maxVec(nums1, i), maxVec(nums2, k - i)));
        }
        return res;
    }
    vector<int> maxVec(vector<int> nums, int k) {
        if (k == 0) return {};
        vector<int> res;
        int drop = nums.size() - k;
        for (auto a : nums) {
            while (drop && res.size() && res.back() < a) {
                res.pop_back();
                --drop;
            }
            res.push_back(a);
        }
        res.resize(k);
        return res;
    }
    vector<int> mergeVec(vector<int> nums1, vector<int> nums2) {
        vector<int> res;
        while (nums1.size() + nums2.size()) {
            vector<int> &t = nums1 > nums2 ? nums1 : nums2;
            res.push_back(t[0]);
            t.erase(t.begin());
        }
        return res;
    }
};