leetcode 33. Search in Rotated Sorted Array 二分查找

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.

就是在一个旋转后的数组中寻找target数据。

很明显，最简单的方法就是遍历一次，不过题目的意思明显不是要求这么做的，主要考察的还是使用二分查找，所以这里使用二分查找来做。

代码如下：

public class Solution {    public int search(int[] nums, int target)     {        if(nums==null || nums.length<=0)            return -1;        /* 下面的属于遍历求解，很明显，不是这么做        int res=-1;        for(int i=0;i<nums.length;i++)        {            if(nums[i]==target)            {                res=i;                break;            }        }        return res;*/        //寻找最小值的index        int minIndex=0;           for(int i=1;i<nums.length;i++)            minIndex=(nums[minIndex]>=nums[i])? i : minIndex;        int res=-1;        //确认是从小到大的排列        if(minIndex==0)            res=binarySearch(nums,0,nums.length-1,target);        else        {            //分成两部分做二分查找            int t=binarySearch(nums,0,minIndex-1,target);            int tt=binarySearch(nums, minIndex, nums.length-1,target);            if(t!=-1)                res=t;            if(tt!=-1)                res=tt;        }        return res;    }    private int binarySearch(int[] nums, int start, int end,int target)    {        if(start>end)            return -1;        else if(start==end)        {            if(nums[start]==target)                return start;            else                return -1;        }else        {            int mid=(start+end)/2;            if(nums[mid] < target)                return binarySearch(nums, mid+1, end, target);            else                return binarySearch(nums, start, mid, target);        }    }}

C++的做法就是直接遍历

代码如下：

#include <iostream>#include <vector>using namespace std;class Solution {public:    int search(vector<int>& nums, int target)     {        for (int i = 0; i < nums.size(); i++)        {            if (nums[i] == target)                return i;        }        return -1;    }};