lc-321


给你两个整数数组 nums1nums2,它们的长度分别为 mn。数组 nums1nums2 分别代表两个数各位上的数字。同时你也会得到一个整数 k

请你利用这两个数组中的数字中创建一个长度为 k <= m + n 的最大数,在这个必须保留来自同一数组的数字的相对顺序。

返回代表答案的长度为 k 的数组。

示例 1:

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

示例 2:

输入:nums1 = [6,7], nums2 = [6,0,4], k = 5
输出:[6,7,6,0,4]

示例 3:

输入:nums1 = [3,9], nums2 = [8,9], k = 3
输出:[9,8,9]

提示:

  • m == nums1.length
  • n == nums2.length
  • 1 <= m, n <= 500
  • 0 <= nums1[i], nums2[i] <= 9
  • 1 <= k <= m + n
class Solution {
    public int[] maxNumber(int[] nums1, int[] nums2, int k) {
        int m = nums1.length, n = nums2.length;
        int[] maxSubsequence = new int[k];
        int start = Math.max(0, k - n), end = Math.min(k, m);
        for (int i = start; i <= end; i++) {
            int[] subsequence1 = maxSubsequence(nums1, i);
            int[] subsequence2 = maxSubsequence(nums2, k - i);
            int[] curMaxSubsequence = merge(subsequence1, subsequence2);
            if (compare(curMaxSubsequence, 0, maxSubsequence, 0) > 0) {
                System.arraycopy(curMaxSubsequence, 0, maxSubsequence, 0, k);
            }
        }
        return maxSubsequence;
    }

    public int[] maxSubsequence(int[] nums, int k) {
        int length = nums.length;
        int[] stack = new int[k];
        int top = -1;
        int remain = length - k;
        for (int i = 0; i < length; i++) {
            int num = nums[i];
            while (top >= 0 && stack[top] < num && remain > 0) {
                top--;
                remain--;
            }
            if (top < k - 1) {
                stack[++top] = num;
            } else {
                remain--;
            }
        }
        return stack;
    }

    public int[] merge(int[] subsequence1, int[] subsequence2) {
        int x = subsequence1.length, y = subsequence2.length;
        if (x == 0) {
            return subsequence2;
        }
        if (y == 0) {
            return subsequence1;
        }
        int mergeLength = x + y;
        int[] merged = new int[mergeLength];
        int index1 = 0, index2 = 0;
        for (int i = 0; i < mergeLength; i++) {
            if (compare(subsequence1, index1, subsequence2, index2) > 0) {
                merged[i] = subsequence1[index1++];
            } else {
                merged[i] = subsequence2[index2++];
            }
        }
        return merged;
    }

    public int compare(int[] subsequence1, int index1, int[] subsequence2, int index2) {
        int x = subsequence1.length, y = subsequence2.length;
        while (index1 < x && index2 < y) {
            int difference = subsequence1[index1] - subsequence2[index2];
            if (difference != 0) {
                return difference;
            }
            index1++;
            index2++;
        }
        return (x - index1) - (y - index2);
    }
}

文章作者: 倪春恩
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