33. Search in Rotated Sorted Array

最新推荐文章于 2020-08-05 08:57:53 发布

原创最新推荐文章于 2020-08-05 08:57:53 发布 · 159 阅读

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本文介绍了一种在旋转排序数组中查找特定元素的有效方法。通过使用改进的二分查找算法，文章提供了一个Python实现示例，并附带了C++代码示例。改进后的算法能够高效地确定中间元素是否位于旋转前或旋转后的部分，并据此缩小搜索范围。

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python写很简单。

class Solution:
    def search(self, nums, target):
        try:
            ans = nums.index(target)
        except:
            ans = -1
        return ans

常规写的话，我看到一个比较好的二分法，就是确定mid的位置是属于rotated前还是后，然后二分逼近。

class Solution:
    def search(self, nums, target):
        beg, end = 0, len(nums)-1
        while beg<=end:
            mid = (beg+end)//2 
            if nums[mid] == target:
                return mid  
            isAscendingLeft = nums[beg] <= nums[mid]
            if isAscendingLeft:
                if nums[beg] <= target < nums[mid]:
                    end = mid-1
                else:
                    beg = mid+1
            else:
                if nums[mid] < target <= nums[end]:
                    beg = mid+1
                else:
                    end = mid-1
              
        return -1

还有c++写法：

class Solution {
public:
    int search(int A[], int n, int target) {
        int lo=0,hi=n-1;
        // find the index of the smallest value using binary search.
        // Loop will terminate since mid < hi, and lo or hi will shrink by at least 1.
        // Proof by contradiction that mid < hi: if mid==hi, then lo==hi and loop would have been terminated.
        while(lo<hi){
            int mid=(lo+hi)/2;
            if(A[mid]>A[hi]) lo=mid+1;
            else hi=mid;
        }
        // lo==hi is the index of the smallest value and also the number of places rotated.
        int rot=lo;
        lo=0;hi=n-1;
        // The usual binary search and accounting for rotation.
        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;
    }
};