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  • 33-Search in Rotated Sorted Array

    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

    我的解:

    Runtime: 4 ms, faster than 80.23% of C++ online submissions for Search in Rotated Sorted Array.
    Memory Usage: 8.8 MB, less than 77.11% of C++ online submissions for Search in Rotated Sorted Array.
    // 二分查找思想,只是二分的条件有所变化
    class
    Solution { public: int search(vector<int>& nums, int target) { int b = 0; int e = nums.size() - 1; while(b <= e) { int mid = b + (e-b)/2; if (nums[mid] == target)return mid; if (nums[mid] < nums[b]) { if (target == nums[e])return e; if (target > nums[mid] && target < nums[e]) b = mid + 1; else e = mid - 1; } else { if (target == nums[b])return b; if (target > nums[b] && target < nums[mid]) e = mid - 1; else b = mid + 1; } } return -1; } };
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  • 原文地址:https://www.cnblogs.com/qiang-wei/p/11801437.html
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