花花酱 LeetCode 321. Create Maximum Number

zxi — Tue, 14 Nov 2017 16:21:18 +0000

Problem:

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.

Example 1:

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

Example 2:

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

Example 3:

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

题目大意：给你两个数字数组和k，返回从两个数组中选取k个数字能够组成的最大值。

Idea: Greedy + DP

Solution:

Time complexity: O(k * (n1+n2)^2)

Space complexity: O(n1+n2)

C++

// Author: Huahua
// Runtime: 26 ms
class Solution {
public:
    vector maxNumber(vector& nums1, vector& nums2, int k) {
        vector ans;
        const int n1 = nums1.size();
        const int n2 = nums2.size();
        for (int i = max(0, k - n2); i <= min(k, n1); ++i)
            ans = max(ans, maxNumber(maxNumber(nums1, i), 
                                     maxNumber(nums2, k - i)));
        return ans;
    }
private:    
    vector maxNumber(const vector& nums, int k) {
        vector ans(k);                
        int j = 0;
        for (int i = 0; i < nums.size(); ++i) {
            while (j > 0 && nums[i] > ans[j - 1] 
                   && nums.size() - i > k - j) --j;
            if (j < k) ans[j++] = nums[i];
        }
        return ans;
    }
    
    vector maxNumber(const vector& nums1, const vector& nums2) {
        vector ans(nums1.size() + nums2.size());
        auto s1 = nums1.cbegin();
        auto e1 = nums1.cend();
        auto s2 = nums2.cbegin();
        auto e2 = nums2.cend();        
        int index = 0;
        while (s1 != e1 || s2 != e2)
            ans[index++] = 
              lexicographical_compare(s1, e1, s2, e2) ? *s2++ : *s1++;
        return ans;
    }
};

Java

// Author: Huahua
// Runtime: 18 ms
class Solution {
    public int[] maxNumber(int[] nums1, int[] nums2, int k) {
        int[] best = new int[0];
        for (int i = Math.max(0, k - nums2.length); 
                 i <= Math.min(k, nums1.length); ++i)            
            best = max(best, 0, 
                       maxNumber(maxNumber(nums1, i), 
                                 maxNumber(nums2, k - i)), 0);
        return best;
    }
    
    private int[] maxNumber(int[] nums, int k) {
        int[] ans = new int[k];
        int j = 0;
        for (int i = 0; i < nums.length; ++i) {
            while (j > 0 && nums[i] > ans[j - 1] 
                && nums.length - i > k - j) --j;
            if (j < k)
                ans[j++] = nums[i];
        }        
        return ans;
    }
    
    private int[] maxNumber(int[] nums1, int[] nums2) {
        int[] ans = new int[nums1.length + nums2.length];
        int s1 = 0;
        int s2 = 0;
        int index = 0;
        while (s1 != nums1.length || s2 != nums2.length)
            ans[index++] = max(nums1, s1, nums2, s2) == nums1 ? 
                           nums1[s1++] : nums2[s2++];
        return ans;
    }
    
    private int[] max(int[] nums1, int s1, int[] nums2, int s2) {
        for (int i = s1; i < nums1.length; ++i) {
            int j = s2 + i - s1;
            if (j >= nums2.length) return nums1;
            if (nums1[i] < nums2[j]) return nums2;
            if (nums1[i] > nums2[j]) return nums1;
        }
        return nums2;
    }
}

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花花酱 LeetCode 321. Create Maximum Number

Idea: Greedy + DP

Solution:

C++

Java