Leetcode: 33. Search in Rotated Sorted Array

Description

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

• 二分查找
• 有两个关键，由于数组经过旋转过，所以应该先判断mid落在前半段还是后半段
• 然后再根据target的大小判断怎么缩小范围

代码

class Solution {
public:
   int search(vector<int>& nums, int target) {
		int len = nums.size();
		if (len == 0) return -1;
		
		int low = 0, high = len - 1, mid = 0;
		while (low <= high){
			mid = low + (high - low) / 2;

			if (nums[mid] == target) return mid;

            //落在前半段
			if (nums[mid] >= nums[low]){
			    //落在前半段的前半段
				if (nums[low] <= target && target < nums[mid])
					high = mid - 1;
				else low = mid + 1;
				
			}
			else{
			    //落在后半段的后半段
				if (nums[mid] < target && target <= nums[high])
					low = mid + 1;
				else high = mid - 1;
			}
		}

		return -1;
	}
};