leetcode-33.Search in Rotated Sorted Array 搜索旋转排序数组

题目：

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

假设按照升序排序的数组在预先未知的某个点上进行了旋转。 ( 例如，数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。 搜索一个给定的目标值，如果数组中存在这个目标值，则返回它的索引，否则返回 -1 。 扫描二维码关注公众号，回复： 4739076 查看本文章 你可以假设数组中不存在重复的元素。 你的算法时间复杂度必须是 O(log n) 级别。 示例 1: 输入: nums = [4,5,6,7,0,1,2], target = 0 输出: 4 示例 2: 输入: nums = [4,5,6,7,0,1,2], target = 3 输出: -1

思路：排序数组首先考虑使用二分查找。两种划分，第一种nums[mid]<nums[j]，说明后面是有序的，看target是否在这一区间，如果在i=mid+1,不在j=mid-1就在左边找。第二种，nums[mid]>num[j],说明mid前面是有序，看target是否在这一区间，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;
    }
};