leetcode-33.Search in Rotated Sorted Array 搜尋旋轉排序陣列

阿新 • • 發佈：2019-01-01

題目：

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

假設按照升序排序的陣列在預先未知的某個點上進行了旋轉。

( 例如，陣列 [0,1,2,4,5,6,7] 可能變為 [4,5,6,7,0,1,2] )。

搜尋一個給定的目標值，如果陣列中存在這個目標值，則返回它的索引，否則返回 -1 。

你可以假設陣列中不存在重複的元素。

你的演算法時間複雜度必須是 O(log n) 級別。

示例 1:

輸入: nums = [4,5,6,7,0,1,2], target = 0
輸出: 4

示例 2:

輸入: nums = [4,5,6,7,0,1,2], target = 3
輸出: -1

思路：排序陣列首先考慮使用二分查詢。兩種劃分，第一種nums[mid]<nums[j]，說明後面是有序的，看target是否在這一區間，如果在i=mid+1,不在j=mid-1就在左邊找。第二種，nums[mid]>num[j],說明mid前面是有序，看target是否在這一區間，如果在j=mid-1,不在i=mid+1

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int i=0,j=nums.size()-1,mid;
        while(i<=j)
        {
            mid = (i+j)/2;
            if(nums[mid]==target) return mid;
            else if(nums[mid]<nums[j])
            {
                if(nums[mid]<target && target<=nums[j]) i=mid+1;
                else j=mid-1;
            }else
            {
                if(nums[i]<=target && target<nums[mid]) j=mid-1;
                else i=mid+1;
            }
        }return -1;
    }
};

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