leetcode 321

问题

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.

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

Example 1:

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

Example 2:

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

Example 3:

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

分析

求解这个问题首先要求解两个子问题。

  1. 给定一个数组 vector<int> nums，从中求出一个长度为 x 的子序列，使其所代表的数字最大。
  2. 给定两个数组，将其按序分配合并，返回最大可能的数组，要保证元素在新数组中相对位置与原数组相同。

NOTE: 这两个子问题的求解都比较典型和重要。merge的时间复杂度是 $O((n+m)^2)$ ，整体的时间复杂度$O(k(n+m)^2)$

代码

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class Solution {
public:
    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {
        int n1 = nums1.size();
        int n2 = nums2.size();
        
        vector<int> ans = {};
        for (int i = 0; i <= k; ++i) {
            if (i > n1 || k-i > n2) continue;
            vector<int> candi = mergeArr(getArr(nums1, i), getArr(nums2, k-i));
            if (candi.size() > ans.size() || 
               (candi.size() == ans.size() && candi > ans))
                ans = candi;
        }
        return ans;
    }
    vector<int> getArr(vector<int> &nums, int x) {
        if (x == 0) return {};
        int to_pop = nums.size() - x;
        vector<int> ret;
        for (int a: nums) {
            while (ret.size() != 0 && ret.back() < a && to_pop > 0) {
                ret.pop_back();
                to_pop--;
            }
            ret.push_back(a);
        }
        return vector<int>(ret.begin(), ret.begin()+x);
    }
    
    bool geater(vector<int> &a, int i, vector<int> &b, int j) {
        while (i < a.size() && j < b.size()) {
            if(a[i] > b[j]) return true;
            if(a[i] < b[j]) return false;
            i++;
            j++;
        }
        if (i == a.size()) return false;
        return true;
    }
    vector<int> mergeArr(vector<int> a, vector<int> b) {
        int n = a.size() + b.size();
        vector<int> ret(n);
        for (int k = 0, i = 0, j = 0; k < n; ++k) {
            if (geater(a, i, b, j))
                ret[k] = a[i++];
            else
                ret[k] = b[j++];
        }
        return ret;
    }
    
};