Problem

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

Solution

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function(nums, target) {
  var len = nums.length
  var left = 0;
  var right = len - 1;
  var mid = 0;

  while (left <= right) {
    mid = left + Math.floor((right - left) / 2);
    if (nums[mid] === target) return mid;
    if (nums[mid] > nums[right]) {
      if (nums[left] <= target && target < nums[mid]) {
        right = mid - 1;
      } else {
        left = mid + 1;
      }
    } else {
      if (nums[mid] < target && target <= nums[right]) {
        left = mid + 1;
      } else {
        right = mid - 1;
      }
    }
  }

  return -1;
};

Explain:

题意：

输入数组是已排序的数组在某个点上翻转了一下，比如 01234 在 2 上翻转，变成 23401，在输入数组里找是否存在某个数

解：二分查找

（规律：输入数组中取任意一段，从中间分开，必定有一边是已排序的）

先通过中间值与末尾值比大小，判断已排序的是左边还是右边，当然也可以通过中间值跟起点值比大小来判断

(目标值等于中间值的话，就结束了，其实也包括二分查找的极限情况，区间里只剩一个值)

然后判断目标值是否在已排序的那边，在的话在这边继续查找，否则去另一半继续查找

Complexity:

• Time complexity : O(log(n)).
• Space complexity : O(1).