There is an integer array nums sorted in ascending order (with distinct values).
Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].
Given the array numsafter the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
You must write an algorithm with O(log n) runtime complexity.
- unique property is that array is sorted in ascending order
- use binary search to look for values that isn't sorted
Solution
// from solutions//https://leetcode.com/problems/search-in-rotated-sorted-array/solutions/273622/javascript-simple-o-log-n-binary-search-solution/
varsearch=function (nums, target) {let left =0;let right =nums.length-1;while (left <= right) {let mid =Math.floor((left + right) /2);if (nums[mid] === target) {return mid; }// When dividing the roated array into two halves, one must be sorted.// Check if the left side is sortedif (nums[left] <= nums[mid]) {if (nums[left] <= target && target <= nums[mid]) {// target is in the left right = mid -1; } else {// target is in the right left = mid +1; } }// Otherwise, the right side is sortedelse {if (nums[mid] <= target && target <= nums[right]) {// target is in the right left = mid +1; } else {// target is in the left right = mid -1; } } }return-1;};