// 540. Single Element in a Sorted Array #include #include #include /* * given a sorted array consisting of only integers where very element appears * exactly twice, except for one element which appears exactly once. return the * single element that appears only once. solution func must be o(ln) time and * o(1) space complexity. */ int singleNonDuplicate(int *nums, int nums_size) { if (nums_size == 1) return nums[0]; if (nums[0] != nums[1]) return nums[0]; if (nums[nums_size - 1] != nums[nums_size - 2]) return nums[nums_size - 1]; int left = 2, right = nums_size - 3, mid; bool ptr_l, ptr_r; while (left < right) { mid = (left + right) / 2; ptr_l = (nums[mid] == nums[mid - 1]) ? true : false; ptr_r = (nums[mid] == nums[mid + 1]) ? true : false; // if ((ptr_l == false) && (ptr_r == false)) if (!ptr_l && !ptr_r) return nums[mid]; if (mid % 2 == 0) { if (ptr_l) right = mid - 2; else left = mid + 2; } else { if (ptr_r) right = mid - 1; else left = mid + 1; } } if (nums[left] != nums[left - 1] || nums[left] != nums[left + 1]) return nums[left]; else return nums[right]; } int main() { int nums1[] = {1, 1, 2, 3, 3, 4, 4, 8, 8}, nums2[] = {3, 3, 7, 7, 10, 11, 11}; printf("%d\n", singleNonDuplicate(nums1, 9)); // expect: 2 printf("%d\n", singleNonDuplicate(nums2, 9)); // expect: 10 }