[leetcode]33. Search in Rotated Sorted Array旋转过有序数组里找目标值

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

题目

一个有序数组，可能进行了循环移位，在里面进行查找。

Solution1: Two Pointers(left&right)

1. We can observe even rotating the array,  at least one part of such array is sorted.

2. Find such sorted array part and do binary search

3. how to find such sorted array?

(1) use two pointers: left, right

(2) if nums[left] < nums[mid],  we can say from left to mid, it is in ascending order, then left part is sorted

(3) if nums[left] > nums[mid], we can say from left to mid, it is NOT in ascending order, then left part is NOT sorted

code

 /*
 Time: O(logn). We use binary search
 Space：O(1) . We only used constant extra space.
 */

 class Solution {
     public int search(int[] nums, int target) {
         // corner case
         if (nums.length == 0) return -1;
         // initialize
         int left = 0;
         int right = nums.length - 1;

         while (left <= right) {
             int mid = left + (right - left) / 2;
             if (nums[mid] == target) return mid;
             // left part is not in ascending order
             else if (nums[mid] < nums[left]) {
                 if (target > nums[mid] && target <= nums[right]) {
                     left = mid + 1;
                 } else {
                     right = mid - 1;
                 }
             }
             // nums[mid] >= nums[left] left side is in ascending order
             else {
                 if (target < nums[mid] && target >= nums[left]) {
                     right = mid - 1;
                 } else {
                     left = mid + 1;
                 }
             }
         }
         return -1;
     }
 }