33. Search in Rotated Sorted Array

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

难度:medium

题目:假设一个按升序排序的数组在某个未知的轴上旋转。

给定一个搜索值,如果能找到则返回其所在数组中的位置,否则返回-1. 假定数组中无重复元素。算法时间复杂度要为O(log n)

思路:二叉搜索

Runtime: 10 ms, faster than 21.83% of Java online submissions for Search in Rotated Sorted Array.

Memory Usage: 26.8 MB, less than 41.44% of Java online submissions for Search in Rotated Sorted Array.

class Solution {
    public int search(int[] nums, int target) {
        int left = 0; 
        int right = nums.length - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (target == nums[mid]) {
                return mid;
            }
            
            // left < mid </> right
            if (nums[left] < nums[mid]) {
                if (target > nums[mid] || target < nums[left]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            } else if (nums[left] > nums[mid]) { // left > mid </> right
                if (target < nums[mid] || target > nums[right]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            } else { // left = mid
                left += 1;
            }
        }
        
        return -1;
    }
}

原文 

https://segmentfault.com/a/1190000018117834

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