[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.

Discussion

这道题可以采用二分查找法。和在一个有序的数组里查找指定元素类似。

如上图所示，有三种情况。

1. nums[leftPos] < nums[rightPos]
2. nums[middlePos] > nums[leftPos]：此时可以判定target是否在middle左边。
3. nums[middlePos] < nums[rightPos]：此时可以判定target是否在middle右边。

算法的时间复杂度为O(logn)。

C++ Code

class Solution {public:    int search(vector<int>& nums, int target) {        if(nums.size() == 0)        {            return -1;        }        return binarySearch(nums, target, 0, nums.size() - 1);    }    //二分查找    int binarySearch(vector<int> & nums, int target, int leftPos, int rightPos)    {        int middlePos = (leftPos + rightPos) / 2;        if(nums[leftPos] == target)        {            return leftPos;        }        if(nums[middlePos] == target)        {            return middlePos;        }        if(nums[rightPos] == target)        {            return rightPos;        }        if(rightPos - leftPos <= 2)        {            return -1;        }        //第一种情况 从左向右递增        if(nums[leftPos] < nums[rightPos])        {            if(target > nums[rightPos] || target < nums[leftPos])            {                return -1;            }            if(target > nums[middlePos])            {                return binarySearch(nums, target, middlePos + 1, rightPos - 1);            }            else            {                return binarySearch(nums, target, leftPos + 1, middlePos - 1);            }        }         //第二种情况 middle左侧递增，右侧两段        if(nums[middlePos] > nums[leftPos])        {            if(target > nums[leftPos] && target < nums[middlePos])            {                return binarySearch(nums, target, leftPos + 1, middlePos - 1);            }            else            {                return binarySearch(nums, target, middlePos + 1, rightPos - 1);            }        }        //第三种情况 middle右侧递增，左侧两端        if(nums[middlePos] < nums[rightPos])        {            if(target > nums[middlePos] && target < nums[rightPos])            {                return binarySearch(nums, target, middlePos + 1, rightPos - 1);            }            else            {                return binarySearch(nums, target, leftPos + 1, middlePos - 1);            }        }        return -1;    }};