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Diffstat (limited to 'search_rotated_array2.cpp')
-rw-r--r-- | search_rotated_array2.cpp | 48 |
1 files changed, 48 insertions, 0 deletions
diff --git a/search_rotated_array2.cpp b/search_rotated_array2.cpp new file mode 100644 index 0000000..0054f04 --- /dev/null +++ b/search_rotated_array2.cpp @@ -0,0 +1,48 @@ +// 81. Search in Rotated Sorted Array II +#include "leetcode.h" + +/* + * there is an integer array 'nums' sorted in non-decreasing order (not + * necessarily with distinct values). before being passed to your function, + * 'nums' is rotated at an unknown pivot index 'k(0 <= k < nums.length)' such + * that the resulting array is '[nums[k], nums[k + 1], ..., nums[n - 1], + * nums[0], nums[1], ..., nums[k - 1]]' (0-indexed). for example + * '[0,1,2,4,4,4,5,6,6,7]' might be rotated at a pivot index 5 and become + * '[4,5,6,6,7,0,1,2,4,4]'. given the array 'nums' after the rotation and an + * integer 'target', return true if 'target' is in 'nums' of 'false' if it is + * not in 'nums'. you must decrease the overall operation steps as much as + * possible. + */ + +class Solution { +public: + bool search(vector<int> &nums, int target) { + int left = 0, right = nums.size() - 1; + while (left <= right) { + int mid = left + (right - left) / 2; + if (nums[mid] == target) + return true; + if (nums[left] == nums[mid] && nums[right] == nums[mid]) + left++, right--; + else if (nums[left] <= nums[mid]) { + if (nums[left] <= target && nums[mid] > target) + right = mid - 1; + else + left = mid + 1; + } else { + if (nums[mid] < target && nums[right] >= target) + left = mid + 1; + else + right = mid - 1; + } + } + return false; + } +}; + +int main() { + Solution obj; + vector<int> nums = {2, 5, 6, 0, 0, 1, 2}; + printf("%d\n", obj.search(nums, 0)); // expect: 1 + printf("%d\n", obj.search(nums, 3)); // expect: 0 +} |