持续创作，加速成长！这是我参与「掘金日新计划 · 6 月更文挑战」的第16天。

题目链接： leetcode.com/problems/se…

1. 题目介绍

There is an integer array nums sorted in ascending order (with distinct values).

【Translate】： 有一个按升序(具有不同的值)排序的整数数组nums。

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].

【Translate】： 在传递给你的函数之前，nums可能是在未知的枢轴索引k（1 <= k <nums.length）上旋转的，因此结果数组为[nums [k]，nums [k+1]，...，nums [n-1]，nums [0]，nums [1]，...，nums [k-1]]（0-索引）。例如，[0,1,2,4,5,6,7]可以在枢轴索引3处旋转，变成[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.

【Translate】： 给定可能的旋转后的数组nums和一个整数目标，如果target在nums中，则返回其索引，如果不在nums中，则返回-1。

You must write an algorithm with O(log n) runtime complexity.

【Translate】： 你必须写一个运行复杂度为O(log n)的算法。

【测试用例】： 【约束】：

2. 题解

2.1 暴力穷举 -- O(n)

  这一题很简单，但是难点在于如何以O(log n)来实现。

    public int search(int[] nums, int target) {
        for (int i = 0; i < nums.length; i++)
        {
            if (nums[i] == target) return i;
        }
        return -1;
    }

2.2 Binary Search -- O(log n)

  nagthane所提供的题解 0ms 100% faster O(log(n)) binary search easy with explaination in comments，和jerry13466所提出的Revised Binary Search几乎一致。

 public int search(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        int l = 0;
        int r = nums.length - 1;
        while(l<r){
            int mid = (l+r)/2;
            if(nums[mid] == target)
                return mid;
            if(nums[l] <= nums[mid]){ //checking if first half array is sorted if so
                if(nums[l] <= target && target < nums[mid]){ //check if target lies in the range if so
                    r = mid - 1;                              // search in first half only
                }else                                         //else search in second half
                    l = mid + 1;
            }else{  //if first half isn't sorted go and check for second
                if(nums[mid] < target && target <= nums[r]){ //check if target lies in second half
                    l = mid + 1;                             //if so search in second half
                }else{
                    r = mid - 1;
                }
            }
        }
        return nums[l] == target ? l : -1;
    }