给你两个整数数组 nums1
和 nums2
,它们的长度分别为 m
和 n
。数组 nums1
和 nums2
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);
}
}