leetcode 33. Search in Rotated Sorted Array (无重复数的旋转数组找定值)

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.

思路：找规律。。。二分。。。

0 1 2 3 4 5 6
1 2 3 4 5 6 0
2 3 4 5 6 0 1
3 4 5 6 0 1 2
4 5 6 0 1 2 3
5 6 0 1 2 3 4
6 0 1 2 3 4 5

观察发现。。。a[mid]<a[r]的时候，后半段有序;

a[mid]>a[r]的时候，前半段有序。。

然后根据有序区间的端点值，确定tar是在该有序区间，还是在另一半区间。

/* ***********************************************
Author :111qqz
Created Time :2017年04月13日 星期四 14时17分03秒
File Name :33.cpp
************************************************ */
class Solution {

public:

    int bin(int n,int tar,vector<int>& nums)
    {
	int l = 0 ;
	int r = n-1;
	while (l<=r)
	{
	    int mid = (l+r)>>1;
	    if (nums[mid]==tar) return mid;
	    if (nums[mid]<nums[r])
	    {
		if (nums[mid]<tar && tar<=nums[r]) l = mid + 1;
		else r = mid -1;
	    }
	    else
	    {
		if (nums[l]<=tar && tar < nums[mid]) r = mid - 1;
		else l = mid + 1;
	    }
	}
	return -1;
    }

    int search(vector<int>& nums, int target) {
    int n = nums.size();
    int res = bin(n,target,nums);
    return res;


    }

};