33. Search in Rotated Sorted Array (opens new window)

Description

Difficulty: Medium

Related Topics: Array (opens new window), Binary Search (opens new window)

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 nums after 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.

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

Example 3:

Input: nums = [1], target = 0
Output: -1

Constraints:

• 1 <= nums.length <= 5000
• -10⁴ <= nums[i] <= 10⁴
• All values of nums are unique.
• nums is an ascending array that is possibly rotated.
• -10⁴ <= target <= 10⁴

Solution

Language: JavaScript

/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 * 1.暴力解法
 * 线性搜索数组
 * 2.二分搜索
 * 关键字：排序，搜索
 * 模式识别：有序或者部分有序，基本使用二分搜索及其变种
 * 算法描述：“丢弃”一半的数据
 * 4 5 6 7 1 2 3 4     target = 5
 * 如果 left < mid 那么左半边有序否则右半边有序
 * 左半边有序时，若目标值再有序区间范围内，
 * 搜索有序侧，否则搜索无序侧
 */
var search = function(nums, target) {
    let [left, right] = [0, nums.length - 1];
    
    while (left <= right) {
        const mid = (left + right) >>> 1;
        if(nums[mid] === target) return mid;
        
        if (nums[mid] > nums[right]) {
            if (nums[left] <= target && nums[mid] > target) right = mid - 1;
            else left = mid + 1;
        }
        
        else {
            if (nums[mid] < target && nums[right] >= target) left = mid + 1;
            else right = mid - 1;
        }
    }
    
    return -1;
};