// 1920. Build Array from Permutation #include "leetcode.h" /* * given a 0-based permutation 'nums', build an array 'ans' of the same length * where 'ans[i] = nums[nums[i]]' for each '0 <= i < nums.len' and return it. a * 0-based permutation 'nums' is an array of distinct integers from 0 to * 'nums.len - 1'. */ int *buildArray(int *nums, int numsSize, int *returnSize) { *returnSize = numsSize; for (int i = 0; i < numsSize; i++) nums[i] += (nums[nums[i]] & 1023) << 10; for (int i = 0; i < numsSize; i++) nums[i] >>= 10; return nums; } int main() { int n1[] = {0, 2, 1, 5, 3, 4}, rs1; int n2[] = {5, 0, 1, 2, 3, 4}, rs2; int *ba1 = buildArray(n1, ARRAY_SIZE(n1), &rs1); int *ba2 = buildArray(n2, ARRAY_SIZE(n2), &rs2); for (int i = 0; i < rs1; i++) printf("%d ", ba1[i]); // expect: 0 1 2 4 5 3 printf("\n"); for (int i = 0; i < rs2; i++) printf("%d ", ba2[i]); // expect: 4 5 0 1 2 3 printf("\n"); }