19.2.2 [LeetCode 33] Search in Rotated Sorted Array

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.

Your algorithm's runtime complexity must be in the order of O(log n).

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

题意

有序数组截成两段打乱顺序（或者也可能保持原序），以logn的复杂度找数

题解

class Solution {
public:
    int search(vector<int>& nums, int target) {
        if (nums.size() <= 0)return -1;
        int pivot = nums[0], size = nums.size(), mididx = size, s=0, e=size-1;
        if (nums[size - 1] <= pivot) {
            int S = 0,E = size - 1;
            while (S <= E) {
                int mid = (S + E) / 2;
                if (nums[mid] >= pivot)
                    S = mid + 1;
                else
                    E = mid - 1;
            }
            mididx = S;
            if (mididx > size - 1)
                mididx = 0;
            if (nums[size - 1] >= target) {
                s = mididx, e = max(size - 1,s);
            }
            else
                s = 0, e = max(mididx - 1,s);
        }
        while (s < e) {
            int mid = (s + e) / 2;
            if (nums[mid] < target)
                s = mid + 1;
            else
                e = mid;
        }
        if (s >= size)return -1;
        if (nums[s] == target)
            return s;
        return -1;
    }
};

先找从哪开始是分界点，然后再找数