g0301_0400.s0321_create_maximum_number.Solution Maven / Gradle / Ivy
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Java-based LeetCode algorithm problem solutions, regularly updated
package g0301_0400.s0321_create_maximum_number;
// #Hard #Greedy #Stack #Monotonic_Stack
public class Solution {
public int[] maxNumber(int[] nums1, int[] nums2, int k) {
if (k == 0) {
return new int[0];
}
int[] maxSubNums1 = new int[k];
int[] maxSubNums2 = new int[k];
int[] res = new int[k];
// select l elements from nums1
for (int l = 0; l <= Math.min(k, nums1.length); l++) {
if (l + nums2.length < k) {
continue;
}
// create maximum number for each array
// nums1: l elements; nums2: k - l elements
maxSubArray(nums1, maxSubNums1, l);
maxSubArray(nums2, maxSubNums2, k - l);
// merge the two maximum numbers
// if get a larger number than res, update res
res = merge(maxSubNums1, maxSubNums2, l, k - l, res);
}
return res;
}
private void maxSubArray(int[] nums, int[] maxSub, int size) {
if (size == 0) {
return;
}
int j = 0;
for (int i = 0; i < nums.length; i++) {
while (j > 0 && nums.length - i + j > size && nums[i] > maxSub[j - 1]) {
j--;
}
if (j < size) {
maxSub[j++] = nums[i];
}
}
}
private int[] merge(int[] maxSub1, int[] maxSub2, int size1, int size2, int[] res) {
int[] merge = new int[res.length];
int i = 0;
int j = 0;
int idx = 0;
boolean equal = true;
while (i < size1 || j < size2) {
if (j >= size2) {
merge[idx] = maxSub1[i++];
} else if (i >= size1) {
merge[idx] = maxSub2[j++];
} else {
int ii = i;
int jj = j;
while (ii < size1 && jj < size2 && maxSub1[ii] == maxSub2[jj]) {
ii++;
jj++;
}
if (ii < size1 && jj < size2) {
if (maxSub1[ii] > maxSub2[jj]) {
merge[idx] = maxSub1[i++];
} else {
merge[idx] = maxSub2[j++];
}
} else if (jj == size2) {
merge[idx] = maxSub1[i++];
} else {
// ii == size1
merge[idx] = maxSub2[j++];
}
}
// break if we already know merge must be < res
if (merge[idx] > res[idx]) {
equal = false;
} else if (equal && merge[idx] < res[idx]) {
break;
}
idx++;
}
// if get a larger number than res, update res
int k = res.length;
if (i == size1 && j == size2 && !equal) {
return merge;
}
if (equal && merge[k - 1] > res[k - 1]) {
return merge;
}
return res;
}
}