【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;
}