Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.
(i.e., [0,1,2,4,5,6,7]
might become [4,5,6,7,0,1,2]
).
You are given a target value to search. If found in the array return its index, otherwise return -1
.
You may assume no duplicate exists in the array.
Your algorithm's runtime complexity must be in the order of O(log n).
Example 1:
Input: nums = [4,5,6,7,0,1,2]
, target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2]
, target = 3
Output: -1
this is my first code, it can't satisfy the last condition:
class Solution { public: int search(vector<int>& nums, int target) { int len = nums.size(); if (len == 0) return -1; bool flag = false; if (target < nums[0]) { for (int i = len-1; i > 0; --i) { if (nums[i] == target) { flag = true; return i; } } } else { for (int i = 0; i < len; ++i) { if (nums[i] == target) { flag = true; return i; } } } if (!flag) { return -1; } } };
this is the right way to use binary search:
class Solution { public: int search(vector<int>& nums, int target) { int len = nums.size(); if (len == 0) return -1; int l = 0; int r = len - 1; while (l <= r) { int m = (l + r) / 2; if (nums[m] == target) return m; if (nums[m] > nums[r]) { if (target < nums[m] && target >= nums[l]) { r = m - 1; } else { l = m + 1; } } else if (nums[m] < nums[l]) { if (target <= nums[r] && target > nums[m]) { l = m + 1; } else { r = m -1; } } else { if (target < nums[m]) { r = m -1; } else { l = m + 1; } } } return -1; } };
Runtime: 8 ms, faster than 25.99% of C++ online submissions for Search in Rotated Sorted Array.
but this way cost more time before the previous way.