【leetcode】Search in Rotated Sorted Array (hard)

Suppose a sorted array 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：二分搜索，每次去掉一半的错误选项。

注意，每次 l = m + 1, r = m - 1 防止无限循环。

int search(int A[], int n, int target) {

    int l = 0, r = n - 1;

    while(l <= r)  //注意有等号

    {

        int m = (l + r) / 2;

        if(A[m] == target)

            return m;

        if(A[l] <= A[m] && A[m] <= A[r]) //顺序的

        {

            if(A[m] > target)

                r = m - 1;

            else

                l = m + 1;

        }

        else if(A[l] >= A[m] && A[m] <= A[r]) //开头转到了左半部分

        {

            if(A[m] < target && target <= A[r]) //在右半部分

                l = m + 1;

            else

                r = m - 1;

        }

        else //开头转到了右半部分

        {

            if(A[l] <= target && target <= A[m]) //在左半部分

                r = m - 1;

            else

                l = m + 1;

        }

    }

    return -1;

}

 

大神简约版写法：去掉一半选项时的思路不同

int search(int A[], int n,int target) {

    int lo = 0;

    int hi = n - 1;

    while (lo <= hi) {

        int mid = (lo + hi) / 2;

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



        if (A[lo] <= A[mid]) {

            if (target >= A[lo] && target < A[mid]) {

                hi = mid - 1;

            } else {

                lo = mid + 1;

            }

        } else {

            if (target > A[mid] && target <= A[hi]) {

                lo = mid + 1;

            } else {

                hi = mid - 1;

            }

        }

    }

    return -1;

}