C语言 | Leetcode C语言题解之第321题拼接最大数
题目：
题解：
int compare(int* subseq1, int subseq1Size, int index1, int* subseq2, int subseq2Size, int index2) {while (index1 < subseq1Size && index2 < subseq2Size) {int difference = subseq1[index1] - subseq2[index2];if (difference != 0) {return difference;}index1++;index2++;}return (subseq1Size - index1) - (subseq2Size - index2);
}int* merge(int* subseq1, int subseq1Size, int* subseq2, int subseq2Size) {if (subseq1Size == 0) {return subseq2;}if (subseq2Size == 0) {return subseq1;}int mergeLength = subseq1Size + subseq2Size;int* merged = malloc(sizeof(int) * (subseq1Size + subseq2Size));int index1 = 0, index2 = 0;for (int i = 0; i < mergeLength; i++) {if (compare(subseq1, subseq1Size, index1, subseq2, subseq2Size, index2) > 0) {merged[i] = subseq1[index1++];} else {merged[i] = subseq2[index2++];}}return merged;
}int* MaxSubsequence(int* nums, int numsSize, int k) {int* stack = malloc(sizeof(int) * k);memset(stack, 0, sizeof(int) * k);int top = -1;int remain = numsSize - k;for (int i = 0; i < numsSize; 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;
}void swap(int** a, int** b) {int* tmp = *a;*a = *b, *b = tmp;
}int* maxNumber(int* nums1, int nums1Size, int* nums2, int nums2Size, int k, int* returnSize) {int* maxSubsequence = malloc(sizeof(int) * k);memset(maxSubsequence, 0, sizeof(int) * k);*returnSize = k;int start = fmax(0, k - nums2Size), end = fmin(k, nums1Size);for (int i = start; i <= end; i++) {int* subseq1 = MaxSubsequence(nums1, nums1Size, i);int* subseq2 = MaxSubsequence(nums2, nums2Size, k - i);int* curMaxSubsequence = merge(subseq1, i, subseq2, k - i);if (compare(curMaxSubsequence, k, 0, maxSubsequence, k, 0) > 0) {swap(&curMaxSubsequence, &maxSubsequence);}}return maxSubsequence;
}