Create Maximum Number

题目地址：
https://leetcode.com/problems/create-maximum-number/description/

题目：

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]

解题思路：
这道题可以用merge 的思想。

代码：

public int[] maxNumber(int[] nums1, int[] nums2, int k) {
    int n = nums1.length, m = nums2.length;
    int[] rst = new int[k];

    // we have Math.max(0, k - m) because if nums2 is shorter than k    // we should starts from k - m from nums1    for(int i = Math.max(0, k - m); i <= k && i <= n; i++){
        int[] candidates = merge(maxArray(nums1, i), maxArray(nums2, k - i), k);
        if(greater(candidates, 0, rst, 0)){
            rst = candidates;
        }
    }
    return rst;
}


// get the max array of number representation with given length - kprivate int[] maxArray(int[] nums, int k){
    int n = nums.length;
    int[] rst = new int[k];
    for(int i = 0, p = 0; i < n; i++){

        // the left number in nums is more than the left space in rst and        // current number in nums is larger than previous value in rst,        // we should move p pointer back by one        while(n - i  > k - p && p > 0 && rst[p - 1] < nums[i]){
            p--;
        }
        if(p < k){
            rst[p++] = nums[i];
        }
    }
    return rst;
}


// merger two array to construct the array with largest valueprivate int[] merge(int[] nums1, int[] nums2, int k){
    int[] rst = new int[k];
    for(int i = 0, j = 0, p = 0; p < k; p++){
        rst[p] = greater(nums1, i, nums2, j) ? nums1[i++] : nums2[j++];
    }
    return rst;
}


// comparing two array with number representation from i and jprivate boolean greater(int[] nums1, int i, int[] nums2, int j){
    while(i < nums1.length && j < nums2.length && nums1[i] == nums2[j]){
        i++;
        j++;
    }

    // condition that the first array's number representation is larger    return j == nums2.length || (i < nums1.length && nums1[i] > nums2[j]);
}