[LeetCode]33. Search in Rotated Sorted Array

题目：Suppose a sorted array is rotatedat some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 01 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 thearray.

翻译

假定一个数组在一个我们预先不知道的轴点旋转。

例如，0 1 2 4 5 6 7可能会变为4 5 6 7 0 1 2。

给你一个目标值去搜索，如果找到了则返回它的索引，否则返回-1。

你可以假定没有重复的元素存在于数组中。

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int l = 0, r = nums.size()-1;
        while (l<=r) {
            int mid = (r-l)/2+l;
            if (nums[mid] == target)
                return mid;
            if (nums[mid] < nums[r]) {
                if (nums[mid]<target && target<=nums[r])
                    l = mid+1;
                else
                    r = mid-1;
            } else {
                if(nums[l]<=target && target<nums[mid])
                    r = mid-1;
                else
                    l = mid+1;
            }
        }
        return -1;
    }
};

法二：

class Solution {
public:
	int search(int A[], int n, int target) {
		int lo = 0, hi = n - 1;
		// 找到最小值的索引
		while (lo<hi){
			int mid = (lo + hi) / 2;
			if (A[mid]>A[hi]) lo = mid + 1;
			else hi = mid;
		}
		// lo==hi是最小值的索引也是数组旋转的位置
		int rot = lo;
		lo = 0; hi = n - 1;
		// 使用二分查找
		while (lo <= hi){
			int mid = (lo + hi) / 2;
			int realmid = (mid + rot) % n;
			if (A[realmid] == target) return realmid;
			if (A[realmid]<target)lo = mid + 1;
			else hi = mid - 1;
		}
		return -1;
	}
};