题解 | #在旋转过的有序数组中寻找目标值#

符合题目时间复杂度要求的二分法求解。

#include<algorithm>
class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param nums int整型vector
     * @param target int整型
     * @return int整型
     */
    int search_fold(vector<int> nums, int left, int right, int target) {
        if (right == left) {
            if (nums[right] == target) {
                return right;
            } else {
                return -1;
            }
        }
        // bipartition the input range, find the ordered range
        int mid = (left + right) / 2;
        int fold_left, fold_right;
        int order_left, order_right;
        if (nums[mid] < nums[left]) {
            fold_left = left;
            fold_right = mid - 1;
            order_left = mid;
            order_right = right;
        } else {
            fold_left = mid + 1;
            fold_right = right;
            order_left = left;
            order_right = mid;
        }
        // firstly search in the ordered range, then search in the folded range
        auto bs_it = lower_bound(nums.begin() + order_left, nums.begin() + order_right,
                                 target);
        if (*bs_it == target) {
            return bs_it - nums.begin();
        } else {
            return search_fold(nums, fold_left, fold_right, target);
        }
    };
    int search(vector<int>& nums, int target) {
        // write code here
        int left(0), right(int(-1) + nums.size());
        return search_fold(nums, left, right, target);
    }
};